One Hundred Problems Involving the Number 100: A Collection of Problems to Celebrate NCTM's First Century 2020022989, 2020022990, 9781680540659, 9781680540666

Citation preview

One Hundred Problems Involving the Number

100 100

A Collection of Problems to Celebrate NCTM's First Century

G. Patrick Vennebush Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

One Hundred Problems Involving the Number 100 A Collection of Problems to Celebrate NCTM’s First Century

G. PATRICK VENNEBUSH

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Copyright © 2020 by The National Council of Teachers of Mathematics, Inc. 1906 Association Drive, Reston, VA 20191-1502 (703) 620-9840; (800) 235-7566; www.nctm.org Library of Congress Cataloging-in-Publication Data Names: Vennebush, G. Patrick, author. Title: One-hundred problems involving the number 100 : celebrate NCTM’s first century / G. Patrick Vennebush. Description: Reston, VA : National Council of Teachers of Mathematics, Inc., [2020] | Summary: “Math educators always seek great problems and tasks for the classroom, and this collection contains many that could be used in various grades. By using this book, the reader will understand ways that great problems can be used to encourage student participation and to promote powerful mathematical ideas. In addition, suggestions for how problems can be presented in the classroom will provide professional development to teachers in the form of effective routines for promoting problem solving. This book would be both a fun read for NCTM’s membership”—Provided by publisher. Identifiers: LCCN 2020022989 (print) | LCCN 2020022990 (ebook) | ISBN 9781680540659 (paperback) | ISBN 9781680540666 (pdf) Subjects: LCSH: Problem solving. | Number theory. | Arithmetic functions. Classification: LCC QA63 .V46 2020 (print) | LCC QA63 (ebook) | DDC 510.71/2—dc23 LC record available at https://lccn.loc.gov/2020022989 LC ebook record available at https://lccn.loc.gov/2020022990 The National Council of Teachers of Mathematics advocates for high-quality mathematics teaching and learning for each and every student. When forms, problems, and sample documents are included or are made available on NCTM’s website, their use is authorized for educational purposes by educators and noncommercial or nonprofit entities that have purchased this book. Except for that use, permission to photocopy or use material electronically from One Hundred Problems Involving the Number 100: A Collection of Problems to Celebrate NCTM’s First Century must be obtained from www.copyright.com, or contact Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. Permission does not automatically extend to any items identified as reprinted by permission of other publishers and copyright holders. Such items must be excluded unless separate permissions are obtained. It will be the responsibility of the user to identify such materials and obtain the permissions. The publications of the National Council of Teachers of Mathematics present a variety of viewpoints. The views expressed or implied in this publication, unless otherwise noted, should not be interpreted as official positions of the Council. Printed in the United States of America

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Table of Contents Acknowledgments..........................................................................................................v Part 1: Why This Book?������������������������������������������������������������������������������������������������ 1

Why 100? ......................................................................................................................................... 1



What Makes a Great Problem? ........................................................................................................ 2



How to Use This Book .................................................................................................................... 6



How This Book Is Organized .......................................................................................................... 7



Problem Solving in the Classroom .................................................................................................. 9



Classroom Practices That Promote a Problem-Solving Culture ................................................. 11



Actions That Inhibit a Problem-Solving Culture ........................................................................ 16

Part 2: The Problems������������������������������������������������������������������������������������������������� 19 Part 3: Solutions and Suggestions��������������������������������������������������������������������������� 41

Problems 1–9................................................................................................................................. 48



Problems 10–19............................................................................................................................. 56



Problems 20–29............................................................................................................................. 65



Problems 30–39............................................................................................................................. 77



Problems 40–49............................................................................................................................. 86



Problems 50–59............................................................................................................................. 96



Problems 60–69...........................................................................................................................107



Problems 70–79...........................................................................................................................123



Problems 80–89...........................................................................................................................137



Problems 90–101.........................................................................................................................152

References���������������������������������������������������������������������������������������������������������������� 171

 iii iii Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Acknowledgments Over the years, I’ve had the privilege of working with myriad wonderful educators, solved many problems from math competitions and conference sessions, and read many amazing books by my peers, so every problem in this book represents an amalgamation of ideas. To say that I’ve been fortunate in my career would be like saying that Euclid was adequate at geometry. I spent several summers at the Johns Hopkins University Center for Talented Youth with a lot of brilliant educators, and countless amazing problems made the rounds among staff. I worked with presidential award winners and lifetime achievement award recipients at the Public Broadcasting Service, Education Testing Service, National Council of Teachers of Mathematics, and Discovery Education. I’ve attended conferences, professional development workshops, webinars, and math circles where great problems were shared on a regular basis. I was the liaison to the Question Writing Committee (QWC) when I worked for the MathCounts Foundation, and later I served as a member of the QWC and eventually as the chair. Which is to say, these problems come from the collective wisdom of hundreds of educators and thousands of resources. Where possible, I give attribution in the text. But to pinpoint the exact source of each problem would be like attempting to identify which butterfly flapped its wings in Madagascar and caused a typhoon in Japan. Instead, I’ll offer gratitude to the math educators who have freely shared their ideas and great problems with me. Through writing, solving, and discussing, my out-of-school education was greatly enhanced by David Barnes, Art Benjamin, Carey Bolster, Judy Ann Brown, Tom Butts, Edward Early, Skip Fennell, Carol Findell, Peg Hartwig, Marjan Hong, Liz Marquez, Harold Reiter, Nicole Rigelman, Jim Rubillo, Marian Small, Dave Sundin, Kris Warloe, and Joshua Zucker. Their influence can be seen throughout the pages of this book. There are hundreds of others who should probably be listed, but the margins are too narrow to contain all their names. I offer a collective thank you to them all. Finally, the completion of this book happened in no small part thanks to the efforts of my family. My wife, Nadine Block, allowed me the time and space to write when I should have been helping with chores or giving her the time and attention she deserves. I am eternally grateful that a wonderful woman was able to find the coping mechanisms necessary to tolerate my shortcomings and idiosyncrasies. My twin sons, Alex and Eli, served as the preliminary editors of this volume, reading every problem multiple times, correcting errors, and suggesting improvements. They are my favorite students, my greatest joy, and the best audience for my terrible jokes, and the quality of this book is a direct result of their tremendous help. I give thanks daily for a family that I love dearly but would never claim to deserve. —G. Patrick Vennebush

 v v Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Part 1 Why This Book? The century of problems in this book range from elementary to advanced. They cover topics from number theory, probability and statistics, geometry, algebra, and operations. Some will inspire you; others will challenge you; but hopefully all of them will intrigue you. Some of them are classics that you’ve seen before; others are brand new, and they are being published for the first time in this volume. All of them have brought joy to me and the students with whom I’ve worked, and I hope that you and your students have a similar experience. In the interest of complete transparency, know that the cover contains a lie. There are actually 101 problems in this book. The extra 1 percent has been provided absolutely free of charge.

WHY 100? Any number that “turns the dial” from a string of 9s to a 1 followed by some 0s is important, but it’s possible that 100 may be the most important power of 10. For elementary teachers and students, the 100th day of school is a big deal, partially because it indicates that the year is more than half over, but also because the number 100 represents a significant milestone in regard to place value. The number 100 is used as the basis for comparisons with percentages. A temperature of 100° C will cause water to boil, there are 100 members of the U.S. Senate, there are 100 years in a century, and there are 100 letter tiles in a Scrabble® game. The 100 most common words in the English language account for more than half of the words used in speaking and writing (Stuart et al. 2003; McNally and Murray 1964). The number 100 is a square number, and it’s the sum of the first nine prime numbers, the sum of the first ten odd numbers, the sum of two square numbers, and the sum of the first four cube numbers: n

100 = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23

n

100 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

n

100 = 64 + 36

n

100 = 64 + 27 + 8 + 1 Why This Book?

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

1

The number 100 is also the sum of several pairs of prime numbers (see problem 44, Prime Time, on page 89). When asked why 100 is special, my 12-year-old son said, “It’s big enough, but it’s not too big” (E. Vennebush, personal communication, April 12, 2020). For instance, counting from 1 to 10 can be done rather quickly, but counting to 100 requires some effort, and getting there feels like an accomplishment. But then continuing to 1,000 or a higher power of 10 would take too long, and just doesn’t seem worth it. That same idea is true beyond counting, too. For patterns and problem solving, the number 100 serves as a good guidepost. Asking students to find the sum of the first 10 positive integers doesn’t feel like much of a challenge. Asking students to find the sum of the first 1,000 positive integers feels like too much of a challenge. But asking students to find the sum of the first 100 positive integers—as Carl Friedrich Gauss was supposedly asked to do by his teacher (see problem 21, Gauss and Check, page 64)—gives the impression that the sum could be calculated by hand or found with a calculator if necessary, but even using a calculator would be tedious enough to suggest that, just maybe, considering a simpler problem or looking for a pattern could be beneficial. The number 100 is just big enough to exhilarate, but not so big as to intimidate. Every problem in this book involves the number 100. Sometimes, a sequence or series will contain 100 terms; other times, the sum of an expression will be 100. Some problems use the number 100 as an exponent, a product, an area, or a perimeter; other problems use 100 as a constant in an equation, as the number of objects in a pile, or as the difference in heights between two dogs. Some problems include a ladder with 100 rungs, a pile with 100 coins, a deck with 100 cards, or a jug that holds 100 ounces. There is something magical about the number 100, and these problems attempt to capture some of that magic. But no matter how 100 is used, each problem is meant to spark curiosity and motivate students (and their teachers) to want to solve it.

WHAT MAKES A GREAT PROBLEM? Consider the following three problems: Problem 1. Let d(n) denote the number of positive divisors of the integer n. Prove that d(n) is odd if and only if n is a square. Problem 2. Which positive integers have an odd number of factors. Justify your answer. Problem 3. Imagine n lockers, all closed, and n men. Suppose the first man goes along and opens every locker. Then the second man goes along and closes every other locker beginning with #2. The third man goes along and changes the state of every third locker beginning with #3 (i.e., if it’s open, he closes it, and vice versa). If this procedure is continued until all n men have passed by all the lockers, which lockers are then open? (Butts 1980, p. 257) 2

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

This book contains 100 problems, and if you’re a math educator, then you’re likely far more interested in those problems than in any information contained in this introduction. I suspect that prior to reading this section, you’ve at least perused the rest of the book, or quite possibly, you’ve attempted every problem contained herein. Consequently, it’s extremely likely that the third problem above looks familiar; it’s very similar to problem 29, The Locker Problem, on page TK. The Locker Problem is a classic that’s been around for quite a while. I first learned of the problem from colleagues in the mid-90s. The above presentation by Butts originally appeared in NCTM’s Forty-Second Yearbook, Problem Solving in School Mathematics, in 1980. I suspect that versions of the problem existed long before that, too. What is it about this problem that has allowed it to survive for more than four decades? Why is this problem and the details of its solution retained by students, while thousands of other traditional textbook exercises are completed and then quickly forgotten? The three problems above are, in fact, the same problem presented in three different ways. That is, their solutions rely on the same underlying mathematics, but the first version states a fact and asks for proof, the second asks a question, while the third offers a novel presentation. Butts (1980) contends, “The phrasing of the third version would significantly motivate the potential solver to tackle the problem” (p. 257). While the format in which the problem is presented is certainly part of its allure, I believe that there are two deeper, more fundamental reasons why the problem appeals to students. First, the mathematics of the Locker Problem is beautifully disguised. Whereas the first two versions use the words divisors and factors, the third version contains very little mathematical language. It isn’t presented as a mathematical problem; it’s just a puzzle. Second, the solution doesn’t rely on the application of previously learned material, but instead it requires students to discover something new about the underlying structure of mathematics. In the vernacular of Common Core, that’s mathematical practice 7. In common parlance, that’s just fun and exciting. The Locker Problem isn’t just an exercise; it’s an actual problem without an obvious solution strategy. The K−8 Publishers’ Criteria (CCSSO 2012) draws a distinction between problems and exercises: “In solving problems, students learn new mathematics, whereas in working exercises, students apply what they have already learned to build mastery” (p. 17). Because students may learn some new mathematics while solving the Locker Problem, that is exactly why it should be used in the classroom. Unfortunately, not everyone holds that belief. I shared the Locker Problem during a conference session recently, and afterwards a middle school teacher told me that she thought it would be a great problem to share with her students after they learned about factors. A similar sentiment was expressed by a student online, claiming that the problem is not appropriate for sixth-grade homework because “there are other ways to teach kids common factorization” (Lollos 2014). Both of those comments unfortunately miss the point and fail to recognize the benefit of great problems. They should not be reserved merely to reinforce Why This Book?

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

3

mathematical skills; they should instead be used to promote the development of conceptual understanding. In the book Switch, Heath and Heath (2010) state, “Script the critical moves. Don’t think big picture, think in terms of specific behaviors” (p. 259). They contend that change happens not when people are asked to make many changes, but when they instead are asked to make just one important change. Using problems at the beginning of lessons is a simple but critical change that could greatly affect mathematics education. Because many great problems already appear in the curriculum, implementing this change would only require a change as to when problems are presented. Rather than assigning them as homework or introducing them after direct instruction, use them instead to open lessons and form the basis for instruction. This shift to a problems-first approach would emphasize problem solving, focus on conceptual development, allow for the natural introduction of vocabulary, and increase learning and retention. Seeley (2014) refers to this as “upside-down teaching” and describes it as having three parts: 1. Students “mess around with a task for a while, ideally engaging in some thinking, trying things out, and generally wrestling with or constructively struggling with mathematics arising from the problem” (p. 90). 2. Then, possible solution strategies are presented in a whole-class discussion, but students will explain, clarify, and continue to struggle with the mathematics. 3. F  inally, the teacher connects the students’ work to big ideas, ensuring that students understand how the problem connects to the important mathematics that they are expected to learn. Note that only in the third part does the teacher have the role of providing any explanation or direct instruction—after students have struggled and presented their work. The Locker Problem, when solved prior to learning about factors, provides an opportunity to introduce terms such as factor, divisor, prime, composite, and square number when they arise organically as part of a class discussion; moreover, it has the potential to unveil deep insights about the structure of our number system and the nature of mathematics: Most integers have an even number of factors because the factors occur in pairs; square numbers have an odd number of factors because the square root of the number occurs as a repeated factor; and the integer 1 is the only number with a single factor. Indeed, when students attempt problems for which they don’t immediately have a solution strategy, they learn more. Brown, Roediger, and McDaniel (2014) note that “Trying to solve a problem before being taught the solution leads to better learning, even when errors are made in the attempt” (p. 4). It is my sincere hope that all of the problems in this book—like the Locker Problem—serve to initiate discovery, exploration, and learning. They are intended to stimulate interest while providing an opportunity for students to better understand some aspect of mathematics. I have always 4

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

contended that great math problems are like good jokes: Both make you think, and the punch line is unexpected. Those are two important elements of a rich mathematical task; the list below provides a more extensive set of criteria.

Criteria of a Great Math Problem n It’s exploratory. That is, the problem requires students to pose conjectures and test them. In

short, students will need to get “messy.” n Students want to get messy with the problem because it’s posed in an interesting way. Often,

they contain real-life contexts—but remember, math is a context! n

The solution strategy isn’t obvious, and there are multiple solution strategies.

n

It feels more like a puzzle than a problem, more like play than work.

n Something—and, hopefully, something mathematical—will be learned by solving the problem. n

It is based on important mathematics.

n The problem is low-floor, high-ceiling, meaning that there are entry points for every student

but also challenges for high-achieving students. n

At least one solution is understandable to every student.

n

 lthough the problem can be solved independently, students will benefit from collaboration A and discussion.

n

The problem provides opportunities for extension.

The Locker Problem meets all of those criteria, including the possibility of an extension:

One day, some students are out sick. Regardless, those present repeat the process and just skip the students who are absent—for instance, if student 3 was absent, then no one would change the state of every third locker. When they finish, only locker #1 is open, and the other lockers are all closed. How many students were absent? (Math Jokes 4 Mathy Folks 2011).

Well, would you look at that? Only TK pages into this book, and already an Easter egg! I hope you had fun solving that extension. The criteria on the list above were generated several years ago, following a problem-solving workshop that I offered to middle school teachers. It was a four-hour session, and the first activity required attendees to put some of their favorite problems onto pieces of chart paper. Those problems were to serve as motivation for the work we’d do for the rest of the day. The instructions specifically Why This Book?

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

5

requested problems that engage students, that instill excitement, and that teachers had used in class with great success. When the dust settled, these were some of the problems that hung on the wall: n

Solve for x: 3x + 7 = 16.

n

What is 40% of 150?

n

Write an expression for six more than a number.

My reaction to the collection was “If these are great problems, then what do bad problems look like?” It’s not that the questions above don’t have a place in the curriculum, but they are exercises, not problems. Further, I don’t suspect that any of them would do much to instill excitement in a middle school student. That experience caused me to generate the Criteria of a Great Math Problem because while I could easily identify problems that weren’t great, I was unable to identify the characteristics of a great problem. The Criteria of a Great Math Problem guided the selection of problems for this book. Does every problem satisfy every criterion? Doubtful. Is every problem in this book great? Maybe not. But hopefully with a deft touch, they can become useful problems for your classroom. Present them as is, or modify them to ensure they are developmentally appropriate, match your instructional goals, and fit your students’ needs.

HOW TO USE THIS BOOK For a math educator, it’s never acceptable to present a problem to students without having first solved the problem yourself. This is a critical step to ensure that the problem is appropriate for students and to identify both common strategies that students will use as well as common mistakes that may occur. Smith and Stein (2011) identified five practices for effectively using problems in the classroom; the five practices are— 1. anticipating likely student responses to challenging mathematical tasks; 2. monitoring students’ actual responses to the tasks (while students work on the tasks in pairs or small groups); 3. selecting particular students to present their mathematical work during the whole-class discussion; 4. sequencing the student responses that will be displayed in a specific order; and 5. connecting different students’ responses and connecting the responses to key mathematical ideas (p. 4). The first practice, anticipating, involves making “an effort to actively envision how students might mathematically approach the instructional task” and “developing considered expectations about 6

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

how students might mathematically interpret a problem, the array of strategies—both correct and incorrect—that they might use to tackle it, and how those strategies and interpretations might relate to the mathematical concepts, representations, procedures, and practices that the teacher would like his or her students to learn” (p. 8). Anticipating, in essence, refers to adequate preparation. Just as the woodsman who was given five minutes to chop down a tree said he’d spend the first two-and-a-half minutes sharpening his axe (Jaccard 1956, p. 12), it would seem wise for a teacher who plans to engage students in an hourlong problem-solving activity to spend half an hour solving the problem in advance. While all five practices identified by Smith and Stein (2011) are important, anticipating is the most critical. Effective classroom use of any task hinges on knowing which strategies and mistakes you might see and how to deal with them when they arise. Anticipating those strategies and mistakes will prepare you to effectively sequence the discussion that follows. An effective technique for anticipating is to refer to published solutions for a problem, such as the solutions found in this book—but first do the problem on your own. Through adequate preparation, teachers are free to attend to the relatively few unexpected responses that might arise in the classroom, to make connections between responses, and to make connections to important mathematical ideas. For example, if you prepare adequately and anticipate three different responses, but then five different strategies surface during class discussions, you’ll only have to improvise on two of them. That means that you’ll be prepared for 60 percent of what occurs in class. Anyone who has ever spent any time in a classroom knows that it’s impossible to be 100 percent prepared, but the further you get from 0 percent, the better. So, to use this book effectively, peruse the problems, identify the ones that you might like to use with your students, and solve them. Then solve them a different way. Is the problem you just solved too difficult for your students? Revise it to make it simpler. Was it too easy? Consider an extension. (The “In the Classroom” section that accompanies each solution in Part 3 provides modifications and extensions for adjusting the level of difficulty.) Still not sure about a problem after modifying or extending? Then don’t use it. There are 99 other problems in this book to choose from. Find a different one. Maybe in a month or two, you can return to a problem that you had to temporarily put aside.

HOW THIS BOOK IS ORGANIZED This book is divided into three parts. In Part 1, Why This Book?, which you are reading now, you’ll find general information about the book, tips for how to use it, and suggestions for how to create a problem-solving culture in your classroom. In Part 2, 100 Problems, you’ll find the treasure trove of gems that were promised on the front cover. All 100 problems contain the number 100. Every problem is accompanied by at least one solution, though it’s certainly possible that any problem could be solved using a multitude of solution strategies. Why This Book?

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

7

The order of the problems was carefully arranged so that the problems roughly increase in difficulty. That is, problem 100 is absolutely more difficult than problem 1. Alternatively, it’s impossible to guarantee that problem n is more difficult than problem n − 1 or even more difficult than problem n − 10. Difficulty is subjective and depends on the solver’s experience. Problems that may seem easy for one student may be difficult for other students, and vice versa. Similar problems have been grouped together. For instance, several problems involving patterns may occur consecutively, followed by a handful of geometry problems. This does not mean, however, that all problems of a given type will occur as a collection. For instance, several straightforward pattern problems (such as those that ask “What do you think comes next?”) appear early in the collection, whereas more difficult pattern problems occur much later, and the two subsets are separated by some algebra, geometry, and number theory problems in between. Every problem is aligned to the Common Core State Standards for Mathematics (CCSSM), and both content and practice standards are identified. But this alignment demands some explanation. The Publishers’ Criteria states that, “Grade-level problems . . . often involve application of knowledge learned in earlier grades. Although students may well have learned this earlier content, they have not learned how it extends to new mathematical situations and applications” (CCSSO 2012, p. 12). Most of the problems in this book are not traditional textbook problems. They are rich mathematical tasks that require perseverance, deep thinking, and application of basic ideas in sometimes novel ways. For instance, problem 20, Farey Tales, page TK, is essentially a problem about ordering fractions, and standard 4.NF.A.2 requires that students be able to “Compare two fractions with different numerators and different denominators” (CCSSI 2010, p. 30), but this problem would be challenging for most fourthgrade students. Instead, it would be more appropriate for middle school students as a group problem-solving activity. Similarly, many problems in this book may be most suitable for students who are one or more grades ahead of the listed standards. Consequently, each problem should be solved in advance by the teacher who plans to use it in the classroom; this is another reminder that educators should engage in the practice of anticipating student responses before presenting a task to students. In Part 3, Solutions and Suggestions, there is a table that identifies a range of grade(s) at which a problem could be used as well as describes the similarities of problems that are grouped together. You’ll find an alignment of each problem to the Common Core State Standards for Mathematics (CCSSM), a description of how the problem might be used in the classroom, suggestions for how to provide assistance to students without divulging the answer or even exposing a solution strategy, and possible extensions. But even with all this advice, how you use these problems and when you use these problems—even if you use these problems—is a decision that you’ll need to make. The appropriateness of any task depends on the developmental readiness of your students, and you know your students better than anyone else. 8

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM SOLVING IN THE CLASSROOM Great problems are a necessary condition for developing a problem-solving culture in the classroom, but not a sufficient condition. In addition to posing problems worth solving, a number of actions will help to facilitate an environment in which students want to engage in the problem-solving process. Your classroom should have the appearance that it’s a problem-solving place. It needs to be a safe space where students can make mistakes, learn from them, and try again. This can be demonstrated with signs and messages that reinforce the problem-solving process and strategies. One of the messages that should be prominently displayed in every classroom would contain the Four Rights of the Learner (Kalinec-Craig 2017): 1. You have the right to be confused. 2. You have the right to claim a mistake. 3. You have the right to speak, listen, and be heard (e.g., engage in conversations, ask questions, share ideas, and listen to the thinking of others). 4. You have the right to write, do, and represent only what makes sense to you (pp. 4−6). In addition, every classroom should contain a poster that describes the problem-solving process, which has been described many times by a variety of educators and mathematicians. One of the best descriptions is also one of the oldest. Pólya (1945) claimed that the problem-solving process consists of four steps: 1. Understand the problem. 2. Devise a plan. 3. Carry out the plan. 4. Look back. If a more contemporary source is desired, the process described by Wolfram (2010) can be displayed instead: 1. Posing the right question 2. Going from the real world to a mathematical model 3. Computing 4. Going from the model back to the real world, to verify Both Pólya’s and Wolfram’s processes are iterative, in that the last step is reflective and may require returning to a previous step or starting over. Similarly, Kalinec-Craig (2017) notes that the Rights Why This Book?

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

9

of the Learner “can help children and teachers embrace productive struggle and mistakes as valuable steps in the process of learning mathematics” (p. 1). When used in tandem, the problem-solving process and the Rights of the Learner send a powerful message to students about the actions, behaviors, and attitudes that are valued in a classroom. To truly establish a culture of problem solving, however, students must be engaged in creating norms for the classroom. One effective technique is to hang a large piece of paper on the wall titled “Problem-Solving Strategies That We Know and Love.” At the beginning of the year, the paper should be blank. When a student demonstrates a particular strategy, it can be added to the list. At first, the teacher will identify new strategies as they are suggested by students. Over time, however, students will come to notice additional strategies and suggest their inclusion. I remember when an excited middle school student noticed a new strategy in my classroom and exclaimed, “Mr. Vennebush! Amanda worked backwards to find that answer! Can we add that to the list?” What kind of strategies should be added to the list? Any strategy that students suggest would be useful. Some strategies will be more sophisticated than others, which is fine. At the end of a typical year, a list might include any or all of the following: n

Trial and check

n

Make a model

n

Use logic

n

Act out the problem

n

Make a conjecture

n

Solve an equation

n Estimate

n

Find a pattern

n

Eliminate possibilities

n

Use a formula

n

Use symmetry

n

Use a rule

n

Consider special cases

n

Use a calculator

n

Wishful thinking

n

Be ingenious!

n Solve a simpler

problem n

Work backwards

n

Use an organized list

n

Make a table

n

Make a chart

n

Make a diagram

n

Draw a picture

n Check for relevant

information n Check for irrelevant

information

Just as a cartoon snowball will gather snow as it rolls downhill, a student’s problem-solving toolbox will accumulate strategies throughout the year, as well as year after year. Rarely, however, will new strategies replace old ones. Instead, new strategies have a cumulative effect, providing students with additional options when problems are encountered. This is an important point because many students view a new mathematics course as a reset. New content often requires new strategies, but previously learned strategies can still be effective! This is especially true when students enter an algebra course. With an overreliance on symbolic manipulation and solving equations, students may abandon working backwards, eliminating possibilities, making a table, drawing a picture, 10

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

solving a simpler problem, or the other strategies that they’ve acquired. Consider problem 57, Absolute Power: If x and y are both integers, how many different solutions exist for |x| + |y| = 100? Defaulting to a symbolic approach yields a system of four equations: 100 = x + y 100 = −x + y 100 = x − y 100 = −x − y While this will work, wouldn’t it be easier to investigate with organized tables? x

y

x

y

x

y

x

y

0

100

0

−100

0

100

0

−100

1

99

1

−99

−1

99

−1

−99

2

98

2

−98

−2

98

−2

−98

3

97

3

−97

−3

97

−3

−97

⋮

⋮

⋮

⋮

⋮

⋮

⋮

⋮

100

0

100

0

−100

0

−100

0

Fundamentally, there is no difference between these two solution strategies. Both will yield that there are 400 different integer solutions to the equation. (Note that there are 101 rows in each of the four tables, but several ordered pairs have been double counted.) Yet the symbolic solution somehow reveals less insight than the organized list. The strategy chart is titled “Problem-Solving Strategies That We Know and Love,” and the word love is included deliberately. Just as true love never ends, effective problem-solving strategies shouldn’t be abandoned when new ones are learned.

CLASSROOM PRACTICES THAT PROMOTE A PROBLEM-SOLVING CULTURE To study mathematics is to solve problems, and the ability to solve problems is the goal of mathematics education. In the first edition of the American Mathematical Monthly (Finkel and Colaw, 1894) wrote, “The solution of problems is one of the lowest forms of mathematical research . . . yet its Why This Book?

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

11

educational value cannot be overestimated. It is the ladder by which the mind ascends into higher fields of original research and investigation. Many dormant minds have been aroused into activity through the mastery of a single problem” (p. 1). There are many actions that teachers can take to promote problem solving and arouse dormant minds in the classroom.

Select problems that are low-floor, high-ceiling. Use rigorous tasks that require a minimum of explanation for students to get started yet provide opportunities for deep mathematical thinking. More important, implement a low-floor, high-ceiling approach in your classroom, which “implies a certain pedagogy” and requires the teacher to believe that every student “can do well in mathematics, regardless of their prior attainment, and making mistakes, struggling, and persevering are all important” (NRICH 2011). Identifying problems from this book that would engage and challenge all of your students could be a good first step.

Engage students in the mathematical practices. The Common Core State Standards for Mathematics describe the Standards for Mathematical Practice as the “varieties of expertise that mathematics educators at all levels should seek to develop in their students” (NGA Center and CCSSO, p. 6). To foster these processes and proficiencies, provide problems that require students to persevere, undertake mathematical reasoning, communicate mathematically, create models, select mathematical tools, be precise, and look for structure and patterns.

Allow students to work together and work alone. There are many benefits of permitting students to work together. Sofroniou and Poutos (2016) note, “Group work permits students to develop a range of critical thinking, analytical, and communication skills; effective teamwork; appreciation and respect for other views, techniques, and problem-solving methods, all of which promote active learning” (p. 1). Further, providing opportunities to collaborate is vital for developing mathematical thinkers and doers who are comfortable analyzing others’ work and accepting criticism from peers. While the benefits of group work are many, there are advantages to allowing students to occasionally work independently, too. For students who are easily distracted, working individually allows for better concentration on the task at hand. Further, students get to make their own decisions—and learn from their mistakes—when working individually. The key is to remember that neither strategy should be used exclusively, and an approach that allows for some individual think time with group work is likely to yield the greatest results.

Be less helpful. It can be difficult to avoid the temptation to offer assistance to students when they get stuck, but perseverating and overcoming adversity is what allows students to feel a sense of accomplishment 12

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

and satisfaction when solving problems. Research has shown that productive struggle is necessary for students to learn mathematics with understanding (Hiebert and Grouws 2007). In a non-mathematical context, Berry (1983) speculated, “It may be that when we no longer know what to do, we have come to our real work, and that when we no longer know which way to go, we have begun our real journey. The mind that is not baffled is not employed. The impeded stream is the one that sings” (p. 97). For the mathematics classroom, Zucker (2012) explained that allowing students to struggle is the very process by which they grow mathematically. “Fighting to get out of its chrysalis helps a butterfly to strengthen its new wings. In fact, were someone to help the butterfly out of its cocoon, bypassing this uncomfortable yet essential struggle, its wings would remain soft and weak, and it would never fly. Similarly, as teachers, we all want to help our students—but, at times, being too helpful is precisely what hurts them most. This is why we must be less helpful” (Zucker 2012, pp. 4−5).

Use a variety of strategies to engage students in problem solving. Leinwand (2015) suggests that there are many strategies for encouraging students to persevere with mathematical tasks, including the use of warm-up questions, gradually revealing information, requiring students to take a guess, or to draw a picture, all before asking students to solve a problem. The following strategies are particularly useful for engaging students in the problem-solving process. n  Problem

Stems. Using a problem stem simply means presenting some or all of the contextual information associated with a problem, without actually stating the problem itself. For instance, problem 3, Letter Product, is presented as follows on page TK: Let A = 1, B = 2, C = 3, . . ., Z = 26, according to their position in the alphabet. The letter product of a word is the product of the values of the letters within the word. For instance, CAT has a letter product of 3 × 1 × 20 = 60. How many common English words have a letter product of 100?

Presenting just the problem stem, however, might mean only revealing the following information to students: Let A = 1, B = 2, C = 3, . . ., Z = 26, according to their position in the alphabet. The letter product of a word is the product of the values of the letters within the word. At that point, it would be possible for students to ask questions, suggest words for which they’d like to find the letter product, and so on. It would also be appropriate to ask students, “What is the letter product of the word CAT?” Although the letter product of CAT is included in the problem statement on page TK, this portion can be omitted in a classroom setting and instead can be converted to a question that promotes understanding, encourages engagement, and allows an entry point for all students. Moreover, for many students omitting the actual question—finding how many words have a letter product of 100—removes the anxiety that often accompanies having to solve an unfamiliar problem. Why This Book?

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

13

n  Three

Reads. The “three reads” strategy literally utilizes three readings of a problem, each time with a different focus. The first time the problem is read, the goal is for students to comprehend the context. The use of a problem stem is actually the first part of the three reads strategy. All numbers and the question itself should be omitted for this first read. During the second read, the goal is for students to understand the mathematical structure of the problem. At this point, numbers can be included, but the question itself should still be omitted. With the third read, students should generate a list of mathematical questions that could be answered. Invariably, students will suggest the very same question that appeared in the original problem statement. If not, and if answering that question is important to meeting particular learning goals, it can be suggested by the teacher and added to the list of other questions.

n  Posing

(or Reposing) Problems. Butts (1980) claimed that posing problems properly is an art; creativity is required to present a problem so that students will be motivated to attempt it. Rather than asking a true-false or multiple-choice question, which could result in regurgitation without understanding, a better option is to ask students to provide an example. For instance, instead of asking, “What is the slope of the line y = 3x + 2?” try “Give an example of a line whose slope is 3.” Providing a numerical example within the problem statement—or better yet, discussing a numerical example with the class before the question is posed—may clarify the question and increase motivation. Finally, pose the problem in such a way that the student is willing to take a guess. Exercises that ask students to prove a statement or show that a theorem is always true decrease motivation. Instead of asking, “Show that no power of 2 can be expressed as a sum of consecutive positive integers,” try “Notice that 3 + 4 + 5 + 6 + 7 = 25 and that 5 + 6 + 7 = 18. Which other numbers can be expressed as a sum of consecutive positive integers? Are there any that can’t?” One of the goals of solving problems is for students to “learn something about the art of problem solving,” and the first step in the process of teaching problem solving “is to pose the problem properly” (Butts 1980, p. 268).

Ask questions. Questions serve a variety of purposes. While some are used to assess understanding or gather information, purposeful questions should be used to probe student thinking; to advance student understanding without taking over or funneling student thinking to a particular procedure or strategy; to elicit reasoning and justification; and, to make the mathematics more visible to students (NCTM 2014, p. 41). Whereas explaining is a practice that keeps the teacher at the center of instruction, effective questioning is critical for student-centered instruction. Reinhart (2000) reminds us, “Never say anything a kid can say! [This goal] has forced me to develop and improve my questioning skills. It also sends a message to students that their participation is essential. Every time I am tempted to tell students something, I try to ask a question instead” (p. 480). Questions can be used to advance student thinking when solving problems. Selecting tasks that are low-floor, high-ceiling will mitigate the possibility that a problem is inaccessible to students, but it may 14

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

still happen sometimes. If it does, you can ask, “What do you notice?” and “What do you wonder?” These questions provide a safe space in which students can share their thoughts without anxiety about having to produce a solution. In general, questions to help students get started should be openended and offer guidance without inferring a solution strategy. Starter questions include “What have you done before that is like this?” or “What happens if you . . .?” Once students have begun, some questions can be used to stimulate thinking and help students to identify patterns and structure, such as “What’s the same? What’s different?” or “What comes next?” or “How could this pattern help you find an answer?” Other questions can be used as formative assessment to provide information to the teacher, but those same questions allow students to analyze their own thinking. They should be asked after students have had sufficient time to make progress, and they can be phrased as “What did you find out?” or “How did you find that out?” or “Why do you think that’s the case?” Finally, discussion questions can be used to promote comparison of strategies after students have worked on a task, and it encourages self-evaluation and reflection. Ask students, “Does anyone have the same solution?” or “Who has a different solution to share?” or “Do you think this is the best solution?”

Encourage guessing. Effective teaching should not consist solely of presenting facts, exhibiting algorithms, and helping students achieve mechanical perfection. That is not mathematics. The process of doing mathematics, according to Lockhart (2002), involves “having ideas, not having ideas, discovering patterns, making conjectures, constructing examples and counterexamples, devising arguments, and critiquing each other’s work” (p. 16). With that as the premise, it is important to remember that the most important aspect of open-ended problems is that they “encourage guessing” (Butts 1980, p. 264), but Butts is referring to thoughtful, educated approximations, not random guesses. An effective strategy for encouraging sophisticated guessing is to ask students to give an estimate that is too high and to give an estimate that is too low (Stadel 2014). The purpose of guessing is to inspire conjectures, which fuel the problem-solving process. For instance, problem 33, At the End, There’s Nothing, asks, “How many zeroes are at the end of 100! when it is computed?” Students might suggest that there will be at least 10 zeroes at the end—there will be a 0 for each decade number in 100!—but there will be fewer than 100 zeroes at the end, because not every factor will contribute a trailing zero, and only 100 will contribute more than one zero. Those estimates provide some boundary on the search. Looking a little further, students may notice that 5! = 120 has one trailing zero, 10! = 3,628,800 has two trailing zeroes, and 15! = 1,307,674,368,000 has three trailing zeroes. Does this pattern continue? This may lead to a conjecture that 100! will have 20 zeroes at the end, 100 since = 20, and that is a worthy conjecture to prove (or at least investigate). 5

Urge students to get messy. To learn, students need opportunities to try, fail, and make corrections. One tactic to facilitate this type of interaction is to supply students with physical materials that allow for adjustments. Why This Book?

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

15

Shah (2020) suggests that all students should work on a marker board. “It turns out that human beings would rather work on erasable surfaces than paper and pencil or pen, because if you get something wrong, it just doesn’t feel as safe as if you do it with a marker board. You can always erase mistakes and fix things.” This recommendation is in alignment with the rough-draft thinking promoted by Jansen et al. (2016−17), in which the class culture reinforced the norms that first attempts were merely to capture thinking and class discussions of rough drafts were to collect more information, not evaluate whether work is right or wrong. Encouraging students to do something—draw a picture, generate some examples, consider a simpler version of the problem, make a guess, anything—is better than doing nothing. Zucker (2017) captured this idea brilliantly with the following advice to students: “Never start by staring at a blank piece of paper. Later, after you have some food for thought, pausing by staring at a non-blank piece of paper might be OK” (p. 2).

ACTIONS THAT INHIBIT A PROBLEM-SOLVING CULTURE Just as there are teacher actions that properly set the stage, the following behaviors should be avoided if a culture of problem solving is the desired outcome.

Permitting students to shout an answer, calling on the same students repeatedly, or allowing answers to be shared without adequate wait time. If your high-achieving students are blurting answers within seconds of you asking a question, it may be an indication that you’re not posing problems with sufficient challenge. Or it may be that the norms in your classroom value speed and don’t inhibit these behaviors. Or perhaps you’re asking low-level, quick-answer questions that require a short response and don’t encourage deeper thought. A classroom culture of problem solving can’t develop if all students are not given sufficient agency and voice.

Imposing a time limit for problem solving. Dead ends and false starts—both of which are useful for learning—may slow the problem-solving process. It’s therefore unwise to assume that a task will be completed in a given amount of time. Moreover, there are times when it’s appropriate for students to put a problem aside, perhaps even permanently, due to frustration, anxiety, fatigue, recognizing that more learning is needed, or simply reaching a point of stalled progress. Champagne (2020) noted, “It’s okay to not finish a math problem,” and Fletcher (2020) furthered this idea by saying, “Even if you don’t finish, you’ll learn something.” Some students may not finish the problems that you ask them to solve, and that’s okay. Are students engaged and making progress? If so, don’t worry about completion, and most important, don’t worry about completion in a predetermined amount of time. 16

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Using low-level, quick-response questions. Questions that rely on recall of facts, information, or procedures rather than requiring reasoning and strategic thinking give the impression that mathematics is about speed and that mathematical thinking should happen quickly. Nothing could be further from the truth! Andrew Wiles took seven years to solve Fermat’s Last Theorem (Singh 2013). When students are stuck, the temptation is to ask low-level questions to get them unstuck, but a more productive approach is to ask broader questions to point them in the right direction. For instance, if students are attempting to find the sum of the first n odd numbers, a low-level question might be “What is 1 + 3? What is 1 + 3 + 5?” but a better question wouldt be “Is there a simpler question that you could solve? Have you ever solved a similar problem? How might that help you here? Is there a pattern?”

Reducing the cognitive load. In an attempt to get students unstuck, it’s not uncommon to ask a question that removes all further mathematical thinking. To find a formula for the sum of the first n odd numbers, students will need to identify a pattern and generalize. Asking, “What is 1 + 3? What is 1 + 3 + 5? What is 1 + 3 + 5 + 7?” reduces a rich problem to a computational exercise. Moreover, drawing the following figure for students to help them “see” what’s happening removes all problem solving.

⋯

⋮

⋱

On the other hand, after students have sufficiently struggled, offered some conjectures, evaluated them, and reached some conclusions, it may then be appropriate to share the figure above to draw a connection between the student’s work and this representation. Delivering effective mathematics instruction requires two parts: (1) employing classroom practices that promote a problem-solving culture and (2) utilizing great problems that encourage students to engage in the mathematical practices. While encouraging a problem-solving culture is primarily your responsibility, this book can provide some great problems to support your efforts. Turn the page to find 100 problems worth solving as well as standards, suggestions for use, answers, and solutions. Enjoy the problems, and may the ideas in this book assist with your efforts to develop effective problem solvers. Why This Book?

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

17

Part 2 The Problems This is the heart of the book—a century of problems involving the number 100. They are presented here without solutions, so you can solve them without bias. Try your best to solve them first before you present them to students. If you need a little help, full solutions are given in Part 3—but give it the old college try before flipping the pages. The problems are arranged roughly in order of difficulty, with the easier problems first, and they are grouped into subsets of similar problems. There is no need to solve them in order, though, so feel free to skip around and try the problems that strike your fancy. To help you search for a particular type of problem, find inspiration for a specific lesson, or pursue a task on a certain topic, the following list of short descriptions will provide some guidance. n

Problems 1 to 3 are based on the linguistic properties of numbers.

n Problems 4 to 7 involve conversions and are provided as straightforward problems that require

an understanding of basic mathematical relationships. n Problems 8 to 12 ask for the placement of digits or operators to create an expression

equal to 100. Because there are many possible combinations, students receive significant computational practice while solving each problem. Problems 13 and 14 are straightforward problems involving the basic operations. Each of them requires computational fluency. n Problems 15 to 20 are classic pattern problems in which the object is to determine the next

number in a sequence. The problems are great candidates for the “What comes next?” classroom routine. n

 roblems 21 to 26 involve sums and series. The first several are classic problems, whereas the P last two are atypical problems for the classroom that require deeper thought.

n Problems 27 to 29 involve factors, though this may not be obvious to students upon initial

inspection. The reliance on factors is disguised in the last problem of this set. n Problems 30 to 41 rely on ideas from number theory, ranging from digits to place value, from

products to exponents. Looking for patterns and applying logic will be needed to solve the problems in this set. The Problems

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

19

n

 roblems 42 and 43 are pattern problems, but both rely on the Fibonacci sequence. This P fact is cleverly hidden in the context of the problem, and it’s possible to solve either problem without recognizing the pattern.

n Problems 44 to 48 return to number theory to investigate sums of prime numbers and

products of positive integers. Problems 49 and 50 push students’ understanding by asking them to investigate properties of numbers in ways they may have never done before. Each problem leads to interesting insights about the nature of the number system. n Problems 51 to 56 are algebra problems at their core, but each can be solved without

resorting to equations and symbolic manipulation. Moreover, the last two problems involve Diophantine equations, so they will defy the typical rules of algebra and will force students to think like a number theorist. n Problems 57 and 58 are not your standard function problems. Each could be solved with

a graphing calculator and inspection, though analysis of the functions themselves will be enough to reveal the solutions. n Problems 59 to 63 involve probability and statistics, but you’ll need more than a pair of dice

and a chi-square test to untangle these conundrums. Each one relies on sophisticated logic, but great satisfaction is realized by solving each of them. n Problems 64 to 72 are geometry problems that involve area, perimeter, volume, and

surface area. Most can be understood by elementary students, but the solutions are far from elementary. They require the use of theorems, formulae, organized lists, and sophisticated logic. n Problems 73 to 77 require knowledge of circles, but the Pythagorean theorem often plays a

role in finding their solutions. n Problems 78 to 83 are fundamentally pattern problems. While 78 is relatively straightforward

and 79 to 81 are inextricably linked, 82 and 83 involve elements of graph theory that provide a context not seen by most K-12 students. n

 roblems 84 to 93 represent the third and final group of pattern problems in the collection. P Although some of them cover traditional topics like the handshake problem and magic squares, many of the others are unlike anything you’ve seen before. The genre of eraseand-replace problems will provide an exercise in perseverance as the patterns are not always easily identified.

n Problems 94 to 100 defy categorization. Their common characteristic is that none of them

has much in common with any other problem in the book. Consequently, they form a unique set of problems to ponder when seeking some outside-the-box investigation. Problem 101 is a bonus logic problem that was too much fun to resist. Have fun! 20

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 1. Spell It Out If all the positive integers from 1 to 100 were spelled out, how many letters would be used? Solution

PROBLEM 2. A Funny Name What name did nine-year-old Milton Sirotta create for the number 10100? Solution

PROBLEM 3. Letter Product Let A = 1, B = 2, C = 3, . . ., Z = 26, according to their position in the alphabet. The letter product of a word is the product of the values of the letters within the word. For instance, CAT has a letter product of 3 × 1 × 20 = 60. How many common English words have a letter product of 100? Solution

PROBLEM 4: It’s Gettin’ Kinda Heavy Which weighs more: $100 worth of quarters or $100 worth of dimes? Solution

PROBLEM 5. Days are Numbered How many days is 100 hours? Solution

PROBLEM 6. Walk It Off How many feet is 100 inches? Solution

PROBLEM 7. Release the Hounds There are 100 puppies to adopt in a shelter, and 99 percent of them are hounds. How many hounds must be adopted from the shelter so that 98 percent of the remaining puppies are hounds? Solution

PROBLEM 8. Digital Throwback Insert operators (+, −, ×, ÷) between the digits 1 to 9 to create a true equation. 1 2 3 4 5 6 7 8 9 = 100 What is the fewest number of operators that must be inserted to create a true equation? Solution The Problems

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

21

PROBLEM 9. Freedom of Expression Place each digit 0 to 9 in one of the blanks to make a true equation. Use each digit exactly once. __ __ − __ __ + __ __ − __ __ + __ __ = 100 Solution

PROBLEM 10. It Doesn’t Add Up Insert only addition (+) symbols to make the following equation true.

Solution

10 9 8 7 6 5 4 3 2 1 = 100

PROBLEM 11. Give and Take Insert only addition (+) and subtraction (−) symbols to make the following equation true. Solution

9 8 7 6 5 4 3 2 1 = 100

PROBLEM 12. What’s It Gonna Take? If you solved Give and Take (problem 11), you know that it’s possible to create an expression equal to 100 using the digits 9 through 1 in descending order. What’s the least positive integer value of n such that the numbers from n to 1, arranged in descending order, will allow you to make an expression equal to 100 that involves only addition and subtraction? Solution

22

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 13. Times Square The multiplication table below contains 100 entries. What is the sum of all entries? ×

1

2

3

4

5

6

7

8

9

10

1

1

2

3

4

5

6

7

8

9

10

2

2

4

6

8

10

12

14

16

18

20

3

3

6

9

12

15

18

21

24

27

30

4

4

8

12

16

20

24

28

32

36

40

5

5

10

15

20

25

30

35

40

45

50

6

6

12

18

24

30

36

42

48

54

60

7

7

14

21

28

35

42

49

56

63

70

8

8

16

24

32

40

48

56

64

72

80

9

9

18

27

36

45

54

63

72

81

90

10

10

20

30

40

50

60

70

80

90

100

Solution

PROBLEM 14. Squares and Square Roots What’s the positive difference between 1002 and 100 ? Solution

PROBLEM 15. What Would Your Computer Think? What do you think comes next? Why? 1, 10, 11, 100, ___ Solution

PROBLEM 16. Desmos Be the Place What do you think comes next? Why? 2, 5, 10, 20, 50, 100, ___ Solution

The Problems

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

23

PROBLEM 17. Even I Know This One What do you think comes next? Why? 4, 16, 36, 64, 100, ___ Solution

PROBLEM 18. Number of Numbers In the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, . . ., what do you think the 100th term would be? Why? Solution

PROBLEM 19. What in the World? In the sequence 0, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, . . ., what do you think the 100th term would be? Why? Solution

PROBLEM 20. Farey Tales A Farey sequence Fn is the set of all fractions from 0 to 1 with every possible denominator less than or equal to n. For instance—  0 1 1 1 2 3 1 F4 =  , , , , , ,   1 4 3 2 3 4 1 What is the 100th term of F100? Solution

PROBLEM 21. Gauss and Check What is 1 + 2 + 3 + ⋯ + 100? Solution

PROBLEM 22. Getting Even

What is 2 + 4 + 6 + ⋯ + 100? Solution

PROBLEM 23. Well, That’s Odd

What is 1 + 3 + 5 + ⋯ + (2 × 100 − 1)? Solution

PROBLEM 24. Square Deal

24

What is the sum of the first 100 square numbers? Solution ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 25. Non-Square Numbers The square numbers are 1, 4, 9, 16, . . ., and the non-square numbers are 2, 3, 5, 6, 7, 8, 10, 11, and so on. What is the 100th non-square number? Solution

PROBLEM 26. In Perfect Harmony 1 1 1 1 1 What integer is closest to + + + +  + ? 1 2 3 4 100 Solution

PROBLEM 27. The Great Divide How many positive integer divisors does 100 have? Solution

PROBLEM 28. Factor Fiction What is the least positive integer with exactly 100 positive integer factors? Solution

PROBLEM 29. The Locker Problem A school with 100 students has 100 lockers numbered 1 to 100. The first student goes along and opens every locker. The second student closes every even-numbered locker. The third student changes the state (that is, opens it if it’s closed, or closes it if it’s open) of every locker whose number is a multiple of 3. The fourth student changes the state of every locker whose number is a multiple of 4. And so on, with the nth student changing the state of every locker whose number is a multiple of n. When all 100 students have finished opening and closing lockers, how many lockers will be open? Which ones? Solution

PROBLEM 30. Zero Hero How many zeroes are at the end of 100100? Solution

PROBLEM 31. Famous Last Digits What are the last two digits of 2100? Solution

PROBLEM 32. More Famous Last Digits What is the units digit of 1100 + 2100 + 3100 + 4100 + 5100 + 6100 + 7100 + 8100 + 9100? Solution The Problems

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

25

PROBLEM 33. At the End, There’s Nothing How many zeroes are at the end of 100! when it’s computed? Solution

PROBLEM 34. Can You Digit? Let S(n) represent the number of digits in integer n. For instance, S(4) = 1 and S(256) = 3. What is S(S(100100))? Solution

PROBLEM 35. Sum Kinda Wonderful The sum of the digits of a number is 100, and none of the digits are 0. What is the largest possible value of the number? Solution

PROBLEM 36. Product Marketing The product of the digits of a number is 100, and none of the digits are 1. What is the largest possible value of the number? Solution

PROBLEM 37. Whole Lotta Nothin’ What is the 100th positive integer that contains a 0? Solution

PROBLEM 38. No Zeroes What is the 100th positive integer that does not contain a 0? Solution

PROBLEM 39. Non-Zero Product Two positive integers have a product of 100, and neither number contains the digit 0. What are the two numbers? Solution

PROBLEM 40. To Say the Least If you write the integers 1 to 100, which digit would appear the least number of times? How many times does it appear? Solution

26

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 41. I Got Chills, They’re Multiplying A number with one hundred 9s (and no other digits) is multiplied by 9. How many 9s are in the resulting product? Solution

PROBLEM 42. Ring, Rang, Rungs A ladder with 100 rungs leads to the top of a tower. You can climb one or two rungs at a time. How many different ways are there to get to the top? Solution

PROBLEM 43. Leo’s Bank Account Leo made a deposit at the bank every day for a week. On Monday and Tuesday, he deposited a positive whole number of dollars, and every day thereafter he deposited an amount equal to the sum of the deposits from the previous two days. On Sunday, he deposited exactly $100. What were the amounts of his deposits on the other days? Solution

PROBLEM 44. Prime Time How many pairs of prime numbers have a sum of 100? Solution

PROBLEM 45. In Their Prime How many sets of three prime numbers have a sum of 100? Solution

PROBLEM 46. Prime Pair Two prime numbers have a sum of 100. What is the maximum possible product? Solution

PROBLEM 47. Max Product Two positive integers have a sum of 100. What is the maximum possible product? Solution

PROBLEM 48. To the Max Some positive integers have a sum of 100. What is the maximum possible product? Solution

The Problems

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

27

PROBLEM 49. Why, Certainly For what value of n can you be certain that n consecutive positive integers have a product that is divisible by 100? Solution

PROBLEM 50. Catching Some Zs For how many positive integer values of n is 100!n an integer? Solution

100

PROBLEM 51. Don’t Put Me on a Shelf How many Styrofoam cups could fit on a shelf 100 inches tall? Solution

PROBLEM 52. Play Ball! Together, a baseball and a bat cost $110. The bat costs $100 more than the baseball. How much did the bat cost? Solution

PROBLEM 53. The Jug Is Half Full A jug full of water weighs 100 ounces. When the jug is half full of water, it only weighs 60 ounces. What is the weight of the empty jug? Solution

PROBLEM 54. Dog Day Afternoon What do you notice? What do you wonder? Can you determine the height of the small dog, the height of the large dog, or the height of the table?

Solution 28

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 55. The Barnyard A farmer buys 100 animals for $100. She buys at least one cow, one pig, and one chicken, but no other type of animal. If a cow costs $10, a pig costs $5, and a chicken costs 50¢, how many of each did she buy? Solution

PROBLEM 56. Sum of Cubes For what integer values of x, y, and z is x3 + y3 + z3 = 100? Solution

PROBLEM 57. Absolute Power If x and y are both integers, how many different solutions exist for |x| + |y| = 100? Solution

PROBLEM 58. Absolutely Dreadful For the function below, what is the least possible value of y? y = |||x − 100| + 100| − 100| + 100 Solution

PROBLEM 59. Don’t Be Mean In a list of 100 numbers, the average value of the first 99 numbers in the list equals the average value of all 100 numbers. What is the 100th number in the list? Solution

PROBLEM 60. Counterintuitive Cards A dealer is holding a deck with 100 cards, 30 of which are red on both sides, 30 of which are blue on both sides, and 40 of which are red on one side and blue on the other. The dealer tells you that he will select a card at random and show you the color on one side of the card. If you correctly predict the color on the other side, he will give you $1. If not, you must give $1 to the dealer. What strategy should you use? Solution

PROBLEM 61. Tickets, Please One hundred passengers board a train with exactly 100 seats. Every person has an assigned seat number, but the first person to board has lost her ticket, so she chooses a seat at random. Every passenger thereafter takes their assigned seat, unless their assigned seat is already occupied, in which case they choose a seat at random. If you are the last passenger to board the train, what’s the probability that your assigned seat is unoccupied? Solution

The Problems

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

29

PROBLEM 62. You Say It’s Your Birthday If there are 100 people in a room, what’s the probability that two of them have the same birthday? Solution

PROBLEM 63. The Envelope, Please You are offered a choice of two envelopes. One of them has twice as much money in it as the other. You select one envelope, open it, and find that it has $100 in it. Should you keep the $100 or should you switch to the other envelope? Solution

PROBLEM 64. Wrecked Angle How many different rectangles with integer side lengths and a perimeter of 100 units are possible? Solution

PROBLEM 65. Area of Influence How many different rectangles with integer side lengths and an area of 100 square units are possible? Solution

PROBLEM 66. Around the Squares The figure below consists of 100 square units with adjacent squares sharing a side. What is the perimeter of the figure?

Solution

30

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 67. Chains of Fools Which of the following chains cannot be extended to have a perimeter of exactly 100 units?

Solution

PROBLEM 68. Rock the Octagon An octagon with side lengths of 9, 10, 11, 12, 13, 14, 15, and 16 units has a perimeter of 100 units. If adjacent sides are perpendicular, what is the maximum possible area of this octagon? Solution

PROBLEM 69. Flattening a Triangle The triangle shown below lies on a flat surface and is pushed at the top vertex. The lengths of the two congruent sides are 100 inches and their length does not change when the top vertex is pushed, but the angle between the two congruent sides will increase, and the base will stretch. What is the maximum possible area of the triangle that can be achieved when the top vertex is pushed? And, what is the length of the base when the maximum area is reached?

Solution

The Problems

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

31

PROBLEM 70. Pieces of Ten The regular decagon below has an area of 100 square units. What is the area of the two shaded regions?

Solution

PROBLEM 71. Prism to the Max How many different rectangular prisms have integer edge lengths and a volume of 100 cubic units? Solution

PROBLEM 72. Run for Cover A rectangular prism with integer edge lengths has a volume of 100 cubic centimeters. What is the maximum possible surface area of this solid? Solution

PROBLEM 73. Strike a Chord Find the area of the annulus if a chord of the outer circle that measures 100 cm is externally tangent to the inner circle.

Solution

PROBLEM 74. In the Distance An historic lighthouse is perched on the cliff of a rocky beach. Standing in the lantern room of the lighthouse, you are approximately 100 feet above the surface of the ocean below. How far can you see to the horizon? Solution 32

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 75. Pipe Dream A meter stick is 100 cm long and is placed inside a pipe. The midpoint of the meter stick is 10 centimeters from the pipe wall. What is the diameter of the pipe?

Solution

PROBLEM 76. Around the World Visualize a piece of wire stretched around the equator. Then, a second piece of wire is wrapped around the Earth at a height of 100 meters above the surface. What is the difference between the length of the wire 100 meters above the Earth’s surface and the length of the wire wrapped around the equator? Solution

PROBLEM 77. The Setting Sun In the figure below, the diameter of the small semicircle is coincident with the diameter of the large semicircle. A line segment parallel to the diameter of the large semicircle is tangent to the small semicircle, and its length is 100 centimeters. What is the area of the shaded region?

Solution

PROBLEM 78. Growing Pattern If the pattern below continues, how many squares would be in Stage 100?

Solution

The Problems

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

33

PROBLEM 79. Survivor, Part 1 There are 100 flags arranged in a circle. You can remove either 1 or 2 flags at a time, and the person to remove the last flag wins. If you go first, how many flags should you remove on your first turn to guarantee that you win? Solution

PROBLEM 80. Survivor, Part 2 Begin with a stack of 100 coins. Two players take turns removing 1, 4, or 8 coins. The player who removes the last coin wins. If both players follow optimal strategy, which player will win, and how many coins should that player take on the first turn? Solution

PROBLEM 81. Survivor, Part 3 One hundred people sitting in a circle are given numbers 1 to 100. Going clockwise around the circle starting with the person holding 1, every second person leaves the circle. As some leave, those remaining form a smaller circle without gaps. They continue to remove people in this manner until only one person remains. What number will be held by the last remaining person? Solution

PROBLEM 82. Complete the Circuit Starting in the top left cell of a 10-by-10 grid of squares, a move can be made to an adjacent cell either vertically or horizontally. A complete circuit consists of starting on the cell numbered 1 and touching every other cell in the grid exactly once. One possible circuit is shown below. What is the sum of all numbers on which a circuit could end?

Solution 34

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 83. Numbering the Vertices As shown in the figure below, it’s possible to create a graph with three vertices so that one vertex has degree 1, one vertex has degree 2, and one vertex has degree 3. (The degree of a vertex is the number of edges connected to that vertex.) It’s also possible to create a graph with 8 vertices so that each degree 1 through 8 is associated with a different vertex. Is it possible to create a graph with 100 vertices so that the vertices have degrees 1 through 100?

Solution

PROBLEM 84. Shake It Off At a party with 100 people, every person shakes hands with every other person. How many total handshakes occur? Solution

PROBLEM 85. Scope It Out 1 1 1 1 + + ++ ? 1× 2 2 × 3 3× 4 99 × 100 Solution

What is

PROBLEM 86. Erase and Replace The numbers 1 to 100 are written on a board. At every stage, two numbers, a and b, are erased from the board and replaced by a + b + ab. For instance, if you erase 14 and 51, you’d replace them with 14 + 51 + 14 × 51 = 779. This process is repeated until only a single number remains. What are the possible values of the last number? Solution

PROBLEM 87. Replace the Hypotenuse Start with the numbers 1 to 100 written on a whiteboard. Erase two numbers and replace them with the hypotenuse of a triangle for which those two numbers are the lengths of the two legs. Continue until just one number remains. What number is it? Solution

PROBLEM 88. An Odd Game The numbers 1 to 100 are written in ascending order on a board. Players take turns placing either + or − between any two numbers. After all the spaces between numbers have been filled The Problems

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

35

with an operator, the result is calculated. If the result is odd, you win. What first move should you make? Solution

PROBLEM 89. Pick ‘n’ Add Anand and Bela are playing a number game. Each of them picks a number from 1 to 10.  (If they pick the same number, they both pick again.) If Anand’s number is smaller, then he gets to add his number to Bela’s, and the result is his new number. But at that point, Bela’s number is now smaller, so she gets to add Anand’s number to her number, and the result is her new number. And so on. For example, if Anand chose 2 and Bela chose 5, then Anand’s new number would be 2 + 5 = 7; Bela’s new number would be 5 + 7 = 12; Anand’s next number would be 7 + 12 = 19; Bela’s next number would be 12 + 19 = 31; and so forth. This continues until one of them reaches 100 or more, and that player wins. What is the best number to choose at the start of this game? Solution

PROBLEM 90. Product of Piles Marty has 100 tokens and divides them into two piles of 40 and 60 tokens. He then records the product, 40 × 60 = 2,400. Then he divides the pile of 40 into two piles of 17 and 23 and records that product, 17 × 23 = 391. He divides the pile of 60 into two piles of 30 and 30 and records the product, 30 × 30 = 900. He continues doing this until he has 100 piles of just one token each. Marty then finds the sum of all 99 products that he recorded. What is the greatest possible value for the sum? Solution

PROBLEM 91. Magic Rectangles A magic rectangle is an m × n array of the positive integers from 1 to m × n such that the numbers in each row have a constant sum and the numbers in each column have a constant sum (although the row sum need not equal the column sum). Shown below is a 3 × 5 magic rectangle with the integers 1 to 15. 6

7

8

9

10

13

3

1

11

12

5

14

15

4

2

How many magic rectangles can be made using the integers 1 to 100? Solution 36

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 92. Fair and Square A magic square is a square grid in which each cell contains a distinct positive integer, and the sum of the integers in each row, column, and diagonal is equal. The sum of the numbers in each row, column, and diagonal is known as the magic sum. Create a 4 × 4 magic square in which the magic sum is 100. Solution

PROBLEM 93. Magic Sum What’s the “magic sum” in a 10 × 10 magic square? Solution

PROBLEM 94. One and Done One hundred tennis players enter a single-elimination tournament. How many matches are needed to determine a champion? Solution

PROBLEM 95. To the Point A dartboard contains regions worth 8 and 13 points. Bart scored exactly 100 points. How many times did he hit the region worth 8 points? Solution

PROBLEM 96. Call to Order All possible permutations of the digits 1, 2, 3, 4, and 5 are written in order from the least to the greatest. What five-digit number is in the 100th position in the list? Solution

PROBLEM 97. Triangles on a Grid What’s the probability that three randomly chosen points on a 10 × 10 lattice will form a triangle?

Solution

The Problems

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

37

PROBLEM 98. Binary Quandary The following grid contains 100 squares. If each square is filled with either a 0 or a 1, in how many different ways can the entire grid be filled so that every row contains an even sum?

Solution

PROBLEM 99. Squares on a Grid How many different squares of any size are contained in the 10 × 10 grid shown below?

Solution

38

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 100. Covering with Squares As shown below, a square grid with 100 smaller squares can be covered by 100 squares (each measuring 1 × 1), by 25 squares (each measuring 2 × 2), or by 13 squares (one 6 × 6, two 4 × 4, two 3 × 3, two 2 × 2, and six 1 × 1).

Find all values of n for which it’s impossible to cover a 10 × 10 grid with n squares of integer side length. Solution

PROBLEM 101. Falsehoods and Fibs Which of the following statements are true? 1. Exactly one of the statements in this list is false. 2. Exactly two of the statements in this list are false. 3. Exactly three of the statements in this list are false. 4. Exactly four of the statements in this list are false. 5. Exactly five of the statements in this list are false.

⋮

100. Exactly 100 of the statements in this list are false.

Solution

The Problems

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

39

Part 3 Solutions and Suggestions This section includes a description of how each problem might be used in the classroom as well as suggestions for how to provide assistance to students without divulging the answer or even exposing a solution strategy. Potential extensions for problems are included whenever possible. You’ll also find an alignment of each problem to the Common Core State Standards for Mathematics (CCSSM). But even with all this advice, how you use these problems and when you use these problems—even if you use these problems—is a decision that you’ll need to make. The appropriateness of any task depends on the developmental readiness of your students, and you know your students better than anyone. The following table identifies a range of grade(s) at which a problem could be used as well as describes the similarities of problems that are grouped together. KEY:

appropriate as challenge problems for students who demonstrate mastery grade-level appropriate appropriate for students of this age, but likely not a significant challenge

Problem

Title

K

1

2

3

4

5

6

7

8

9

10 11 12

CCSSM

Problems 1–3 are based on the linguistic properties of numbers. 3.NBT.A.2, 4.NBT.B.5, MP.7

1

Spell It Out

2

A Funny Name

8.EE.A.1, 8.EE.A.3

3

Letter Product

4.OA.B.4, MP.3 

Problems 4–7 involve conversions and are provided as straightforward problems that require an understanding of basic mathematical relationships. 4

It’s Gettin’ Kinda Heavy

4.MD.A.2, 6.NS.B.3, MP.6 (Continued ) Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

41

Problem

Title

K

1

2

3

4

5

6

7

8

9

10 11 12

CCSSM

5

Days are Numbered

4.MD.A.2, 5.MD.A.1, MP.6

6

Walk It Off

4.MD.A.2, 5.MD.A.1, MP.6

7

Release the Hounds

6.RP.A.3.C, 7.RP.A.3, MP.1, MP.6

Problems 8–12 ask for the placement of digits or operators to create an expression equal to 100. Because there are many possible combinations, students receive significant computational practice while solving each problem. Problems 13 and 14 are straightforward problems involving the basic operations. Each of them requires computational fluency. 8

Digital Throwback

2.NBT.B.5, 2.NBT.B.7, 3.NBT.A.2, MP.3

9

Freedom of Expression

3.NBT.A.2, 4.NBT.A.1, MP.2, MP.6

10

It Doesn't Add Up

2.NBT.B.5, 3.NBT.A.2, MP.1, MP.7

11

Give and Take

2.NBT.B.5, 2.NBT.B.7, 3.NBT.A.2, MP.2

12

What's It Gonna Take?

2.NBT.B.5, 2.NBT.B.7, 3.NBT.A.2, MP.3

13

Times Square

4.OA.B.4, 4.NBT.B.5, MP.7

14

Squares and Square Roots

8.EE.A.2, MP.6

Problems 15–20 are classic pattern problems in which the object is to determine the next number in a sequence. The problems are great candidates for the "What comes next?" classroom routine.

42

15

What Would Your Computer Think?

16

Desmos Be the Place

4.OA.C.5, HSF.BF.A.1, MP.7

17

Even I Know This One

4.OA.C.5, 8.EE.A.2, HSF.BF.A.1, MP.3

18

Number of Numbers

4.OA.C.5, 5.OA.B.3, HSF.BF.A.1.A, MP.8

19

What in the World?

4.OA.B.4, 4.OA.C.5, MP.7

20

Farey Tales

4.OA.C.5, 4.NBT.A.2, MP.7

4.OA.C.5, 4.NF.A.2, 6.RP.A.3, MP.7 (Continued )

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Problem

Title

K

1

2

3

4

5

6

7

8

9

10 11 12

CCSSM

Problems 21–26 involve sums and series. The first several are classic problems, whereas the last two are atypical problems for the classroom that require deeper thought. 3.NBT.A.2, HSA.SSE.A.2, MP.7

21

Gauss and Check

22

Getting Even

23

Well, That's Odd

3.NBT.A.2, HSA.SSE.A.2, MP.7

24

Square Deal

HSA.REI.C.6, HSA.REI.C.8, HSF.BF.A.1, MP.7

25

Non-Square Numbers

4.NBT.B.5, 8.EE.A.2, MP.7

26

In Perfect Harmony

3.OA.B.5, 3.NBT.A.2, HSA.SSE.A.2, MP.7

5.NF.A.1, MP.5

Problems 27–29 involve factors, though this may not be obvious to students upon initial inspection. The reliance on factors is disguised in the last problem of this set. 27

The Great Divide

4.OA.B.4, 4.NBT.B.6, MP.7

28

Factor Fiction

4.OA.B.4, 4.NBT.B.6, MP.7

29

The Locker Problem

4.OA.B.4, MP.4

Problems 30–41 rely on ideas from number theory, ranging from digits to place value, from products to exponents. Looking for patterns and applying logic will be needed to solve the problems in this set. 30

Zero Hero

4.NBT.A.2, 8.EE.A.1, 8.EE.A.3, MP.1

31

Famous Last Digits

4.OA.C.5, 8.EE.A.1, MP.7, MP.8

32

More Famous Last Digits

8.EE.A.1, MP.7, MP.8

33

At the End, There’s Nothing

4.OA.C.5, 5.NF.A.1, HSS.CP.B.9, MP.8

34

Can You Digit?

8.F.A.1, HSF.IF.A.2, MP.2

35

Sum Kinda Wonderful

36

Product Marketing

37

Whole Lotta Nothin'

2.NBT.A.2, 4.NBT.A.1, MP.8

38

No Zeroes

2.NBT.A.2, 4.NBT.A.1, MP.8

4.NBT.A.1, 4.NBT.A.2, MP.3 4.NBT.A.2, 4.OA.B.4, MP.3

(Continued ) Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

43

Problem

Title

K

1

2

3

4

5

6

7

8

9

10 11 12

CCSSM

39

Non-Zero Product

40

To Say the Least

2.NBT.A.1, 2.NBT.A.2, MP.7, MP.8

41

I Got Chills, They’re Multiplying

4.NBT.A.2, 5.NBT.A.3.A, 5.NBT.B.5, MP.7, MP.8

4.OA.B.4, 4.NBT.B.5, MP.6

Problems 42 and 43 are pattern problems, but both rely on the Fibonacci sequence. This fact is cleverly hidden in the context of the problem, and it’s possible to solve either problem without recognizing the pattern. 42

Ring, Rang, Rungs

4.OA.C.5, MP.8

43

Leo's Bank Account

4.NBT.B.4, 7.EE.B.4, MP.5, MP.8

Problems 44–48 return to number theory to investigate sums of prime numbers and products of positive integers. Problems 49 and 50 push students' understanding by asking them to investigate properties of numbers in ways they may have never done before. Each problem leads to interesting insights about the nature of the number system. 44

Prime Time

4.OA.B.4, MP.6

45

In Their Prime

4.OA.B.4, MP.6

46

Prime Pair

47

Max Product

3.NBT.A.2, 5.NBT.B.5, MP.6

48

To The Max

3.NBT.A.2, 5.NBT.B.5, 8.EE.A.1, MP.6

49

Why, Certainly

4.OA.A.3, HSA.APR.A.1, HSA.APR.C.4, MP.3, MP.8

50

Catching Some Zs

4.OA.B.4, 5.NBT.B.5, MP.6

8.EE.A.1, HSS.CP.B.9, MP.3, MP.6

Problems 51–56 are algebra problems at their core, but each can be solved without resorting to equations and symbolic manipulation. Moreover, the last two problems involve Diophantine equations, so they will defy the typical rules of algebra and will force students to think like a number theorist.

44

51

Don't Put Me on a Shelf

52

Play Ball!

2.MD.A.1, 4.MD.A.2, 8.EE.C.7, MP.4 8.EE.C.8, MP.2 (Continued )

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Problem

Title

K

1

2

3

4

5

6

7

8

9

10 11 12

CCSSM

53

The Jug Is Half Full

54

Dog Day Afternoon

8.EE.C.8.C, HSA.CED.A.1, MP.4

55

The Barnyard

7.EE.B.3, HSA.CED.A.2, MP.2

56

Sum of Cubes

HSA.SSE.A.1.B, HSN. RN.A.2, MP.2

8.EE.C.8, MP.2

Problems 57 and 58 are not your standard function problems. Each could be solved with a graphing calculator and inspection, though analysis of the functions themselves will be enough to reveal the solutions. 57

Absolute Power

58

Absolutely Dreadful

6.NS.C.7.C, HSA.REI.D.11, HSF.IF.C.7, MP.5 6.NS.C.7.C, 7.NS.A.1.C, HSA.REI.D.11, MP.7

Problems 59–63 involve probability and statistics, but you’ll need more than a pair of dice and a chi-square test to detangle these conundrums. Each one relies on sophisticated logic, but great satisfaction is realized by solving each of them. 6.SP.A.3, HSS.ID.A.2, MP.1, MP.5

59

Don't Be Mean

60

Counterintuitive Cards

61

Tickets, Please

7.SP.C.6, HSS.MD.B.7, MP.3

62

You Say It's Your Birthday

7.SP.C.8, HSS.CP.A.1, MP.4

63

The Envelope, Please

7.SP.C.6, HSS.CP.A.3, HSS.IC.B.5, MP.4

7.SP.A.2, 7.SP.C.6, HSS.MD.B.5.A, MP.3

Problems 64–72 are geometry problems that involve area, perimeter, volume, and surface area. Most can be understood by elementary students, but the solutions are far from elementary. They require the use of theorems, formulae, organized lists, and sophisticated logic. 64

Wrecked Angle

4.MD.A.3, MP.8

65

Area of Influence

4.MD.A.3, MP.8

66

Around the Squares

3.MD.D.8, MP.3

67

Chains of Fools

68

Rock the Octagon

4.G.A.2, 5.OA.B.3, 8.F.A.3, 8.F.B.4, MP.4 3.MD.D.8, 4.MD.A.3, 5.G.B.3, MP.4 (Continued ) Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

45

Problem

Title

K

1

2

3

4

5

6

7

8

9

10 11 12

CCSSM

69

Flattening a Triangle

6.G.A.1, HSN.Q.A.3, HSG.MG.A.3, MP.5

70

Pieces of Ten

HSG.SRT.B.5, MP.6

71

Prism to the Max

5.MD.C.3.B, 5.MD.C.5.B, 6.G.A.2, 7.G.B.6, MP.3

72

Run for Cover

7.G.B.6, MP.4

Problems 73–77 require knowledge of circles, but the Pythagorean theorem often plays a role in finding their solutions. 73

Strike a Chord

7.G.B.4, HSG.C.A.2, MP.3, MP.4

74

In the Distance

8.G.B.7, HSG.C.A.2, MP.4

75

Pipe Dream

8.G.B.7, HSG.C.A.2, MP.4

76

Around the World

77

The Setting Sun

7.G.B.4, MP.4 7.G.B.4, HSG.C.A.2, MP.2

Problems 78–83 are fundamentally pattern problems. While 78 is relatively straightforward and 79-81 are inextricably linked, 82 and 83 involve elements of graph theory that provide a context not seen by most K-12 students. 78

Growing Pattern

5.OA.A.2, 7.EE.A.2, HSA.SSE.A.1

79

Survivor, Part 1

1.OA.C.6, MP.7

80

Survivor, Part 2

2.OA.A.1, 4.OA.C.5, MP.3

81

Survivor, Part 3

4.OA.C.5, HSF.IF.A.3, MP.3

82

Complete the Circuit

4.OA.C.5, MP.3, MP.8

83

Numbering the Vertices

4.OA.C.5, MP.7, MP.8

Problems 84–93 represent the third and final group of pattern problems in the collection. Although some of them cover traditional topics like the handshake problem and magic squares, many of the others are unlike anything you’ve seen before. The genre of erase-and-replace problems will provide an exercise in perseverance as the patterns are not always easily identified.

46

84

Shake It Off

4.OA.C.5, MP.1, MP.4

85

Scope It Out

4.OA.C.5, 5.NF.A.1, HSS.CP.B.9, MP.8

86

Erase and Replace

HSF.IF.A.2, HSF.BF.A.1.A, HSS.CP.B.9, MP.3 (Continued )

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Problem

Title

K

1

2

3

4

5

6

7

8

9

10 11 12

CCSSM

87

Replace the Hypotenuse

88

An Odd Game

89

Pick 'n Add

90

Product of Piles

4.OA.C.5, 4.NBT.B.5, MP.8

91

Magic Rectangles

3.NBT.A.2, 4.NBT.B.5, MP.1

92

Fair and Square

2.NBT.B.6, 3.NBT.A.2, MP.8

93

Magic Sum

3.NBT.A.2, 6.SP.B.5.C, , MP.1

8.G.B.7, HSG.SRT.B.4, HSF.BF.A.1, MP.7 4.OA.C.5, MP.3, MP.7 3.NBT.A.2, MP.8

Problems 94–100 defy categorization. Their common characteristic is that none of them has much in common with any other problem in the book. Consequently, they form a unique set of problems to ponder when seeking some outside-the-box investigation. Problem 101 is a bonus logic problem that was too fun to resist. 94

One and Done

3.OA.A.3, 3.OA.C.7, MP.1

95

To the Point

3.OA.A.3, 8.EE.C.7, MP.4

96

Call to Order

4.NBT.A.2, 4.OA.C.5, HSS.CP.B.9, MP.7, MP.8

97

Triangles on a Grid

8.F.A.3, HSS.CP.B.9, MP.6

98

Binary Quandary

HSS.CP.B.9, MP.1, MP.7

99

Squares on a Grid

5.G.B.3, 6.G.A.3, MP.7

100

Covering with Squares

6.G.A.1, 8.EE.A.2, HSG.MG.A.3, MP.7

101

Falsehoods and Fibs

MP.3

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

47

PROBLEM 1. Spell It Out Common Core State Standards for Mathematics: 3.NBT.A.2 • 4.NBT.B.5 • MP.7

In the Classroom Although the CCSSM standards listed above are for third and fourth grade, this problem could be used with students from first grade through middle school and into high school. The solution given below doesn’t require that the number of letters in each word be counted; instead, some efficiency can be found by noticing patterns. Using those patterns to determine the total number of letters might involve multiplying two-digit numbers, which is a grade 4 standard. But it’s also possible for younger students to find the total using only addition, and it may be that older students would notice other aspects that allow for even more elegant solutions. Problems that use words and letters as a context often engage students who prefer language and humanities to mathematics. As such, these types of problems are exceptional for use at the beginning of the year when attempting to establish a classroom culture. They can set the stage for the type of exploratory problems that form the foundation of an inquiry-based classroom. Math problems about language are plentiful, and the following questions provide possible extensions: n

 hat is the next letter in the sequence O, T, T, F, F, S, S, __? (E: The sequence is formed by W the first letters of the whole numbers 1 to 7.)

n

What is the first number in which the letter a appears when spelled out? (1,000)

n What is the smallest number that, when spelled out, has its letters in alphabetical order?

(40: In fact, it is the only number whose letters are in alphabetical order.) n

 ow many different numbers can be spelled using some of the letters in interchangeability? H (11: The numbers are three, eight, nine, ten, thirteen, thirty, thirty-nine, eighty, eightynine, ninety, and ninety-eight. In fact, interchangeability contains the most number words of any English word.)

Answer and Solution 864. In every group of ten—except for the teens—the numbers one, two, three, four, five, six, seven, eight, and nine are repeated. Those nine words account for 36 letters. Those words are paired with twenty, thirty, . . ., ninety, and each of those prefixes is used ten times. Using these facts saves a lot of time compared to counting the letters. All that is then required is to manually count the number of letters in the teens and also account for the number one hundred. In total, there are 864 letters. 48

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

The following table shows the numbers 1 to 100 and the number of letters in each number. Note that the hyphen is not counted as a letter. Number

Written Out

Letters

Number

Written Out

Letters

1

one

3

25

twenty-five

10

2

two

3

26

twenty-six

9

3

three

5

27

twenty-seven

11

4

four

4

28

twenty-eight

11

5

five

4

29

twenty-nine

10

6

six

3

30

thirty

6

7

seven

5

31

thirty-one

9

8

eight

5

32

thirty-two

9

9

nine

4

33

thirty-three

11

10

ten

3

34

thirty-four

10

11

eleven

6

35

thirty-five

10

12

twelve

6

36

thirty-six

9

13

thirteen

8

37

thirty-seven

11

14

fourteen

8

38

thirty-eight

11

15

fifteen

7

39

thirty-nine

10

16

sixteen

7

40

forty

5

17

seventeen

9

41

forty-one

8

18

eighteen

8

42

forty-two

8

19

nineteen

8

43

forty-three

10

20

twenty

6

44

forty-four

9

21

twenty-one

9

45

forty-five

9

22

twenty-two

9

46

forty-six

8

23

twenty-three

11

47

forty-seven

10

24

twenty-four

10

48

forty-eight

10

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

49

50

Number

Written Out

Letters

Number

Written Out

Letters

49

forty-nine

9

75

seventy-five

11

50

fifty

5

76

seventy-six

10

51

fifty-one

8

77

seventy-seven

12

52

fifty-two

8

78

seventy-eight

12

53

fifty-three

10

79

seventy-nine

11

54

fifty-four

9

80

eighty

6

55

fifty-five

9

81

eighty-one

9

56

fifty-six

8

82

eighty-two

9

57

fifty-seven

10

83

eighty-three

11

58

fifty-eight

10

84

eighty-four

10

59

fifty-nine

9

85

eighty-five

10

60

sixty

5

86

eighty-six

9

61

sixty-one

8

87

eighty-seven

11

62

sixty-two

8

88

eighty-eight

11

63

sixty-three

10

89

eighty-nine

10

64

sixty-four

9

90

ninety

6

65

sixty-five

9

91

ninety-one

9

66

sixty-six

8

92

ninety-two

9

67

sixty-seven

10

93

ninety-three

11

68

sixty-eight

10

94

ninety-four

10

69

sixty-nine

9

95

ninety-five

10

70

seventy

7

96

ninety-six

9

71

seventy-one

10

97

ninety-seven

11

72

seventy-two

10

98

ninety-eight

11

73

seventy-three

12

99

ninety-nine

10

74

seventy-four

11

100

one hundred

10

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 2. A Funny Name Common Core State Standards for Mathematics: 8.EE.A.1 • 8.EE.A.3

In the Classroom Big numbers fascinate students. Mathematician Edward Kasner used this fact to his advantage, and he sought to pique the interest of young students by asking them to suggest a name for the numeral 1 followed by 100 zeroes. The young Milton Sirotta was the nephew of Kasner, and he suggested the name googol on a walk with his uncle. In their book Mathematics and the Imagination, Kasner and Newman (1940) told the story of young Milton coining the term, and googol has been used ever since. This problem can serve as the basis for a lesson in obscure math terms. Students can explore the terms googolplex, octothorpe, vinculum, virgule, anthyphairetic ratio, surd, Jacob’s staff, apothem, Torricelli’s trumpet, oblate spheroid, brute force, the Witch of Agnesi, and even the Hairy Ball theorem (if your students are mature enough).

Answer and Solution Googol. When written out, 10100 is a 1 followed by 100 zeroes. The term was coined in 1938, and it has been in use ever since. In fact, the deliberate misspelling as Google is the name of the tech company that specializes in Internet-related services and products.

PROBLEM 3. Letter Product Common Core State Standards for Mathematics: 4.OA.B.4 • MP.3

In the Classroom To get students invested in the problem, use just the stem. State that the letter product of cat is 60, and then ask them, “What do you notice? What do you wonder?” Invariably, students will deduce that letter product must somehow refer to the product of the letters; if they don’t make the deduction at first, you can provide additional examples with the letter products of other small words, such as be (10) and in (126). Further, students will start to wonder about the letter product of other words, including their names. They may also wonder if any other words have the same letter product as one of the words already presented. This is a good opportunity to pose the following question: What words besides cat have a letter product of 60? Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

51

Some students will immediately realize that act is an anagram of cat and will have the same letter product. Other students will find ale, bob, and do, and possibly more. Students love to make this problem personal and will immerse themselves in the following task: Find a common English word with the same letter product as your name. Be forewarned that this puzzle is not meant for Christopher (255,301,632,000), Xenophon (677,376,000), or Baghyawati (11,592,000)!

Answer and Solution There are nine words with a product value of 100: ate, baby, day, dee, eat, et, eta, tae, and tea. (Note that not all of these nine words will be found in standard dictionaries. All are listed in ENABLE, the Enhanced North American Benchmark LExicon, a definitive collection of more than 172,000 words.) The prime factorization of 100 is 22 × 52, which means the word could contain any of the sets of letters {B, B, E, E}, {B, B, Y}, {B, E, J}, {E, T}, {D, Y}, {D, E, E}, {J, J}. Of course, since its value is 1, any number of As could be appended to any of these sets. Some of the sets seem unlikely to lead anywhere, such as {J, J}, but others hold promise, such as {E, T}, which leads to six words: et, ate, eat, eta, tae, and tea. From there, it becomes a matter of playing with combinations to determine the nine English words that can be formed.

PROBLEM 4. It’s Gettin’ Kinda Heavy Common Core State Standards for Mathematics: 4.MD.A.2 • 6.NS.B.3 • MP.6

In the Classroom It’s absolutely possible to solve this problem by searching online for the weights of coins, which can be found at the U.S. Mint’s website: https://www.usmint.gov/learn. But the problem could be so much richer, touching on a variety of standards, by allowing students to explore with real coins and a balance scale. It would also provide an opportunity for students to consider a fundamental problem-solving strategy: solve a simpler problem. Weighing $100 worth of quarters or dimes could be problematic, not to mention that even a classroom full of students likely doesn’t have that kind of money. But weighing just $1 of each coin is possible and could reveal some insights. The problem could also provide the impetus for a measurement lesson in which the focus is on precision. Coins could be measured using a scale, but there are several considerations to ensure accurate results. First, several coins should be weighed and the average should be found as not all coins 52

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

of the same denomination will weigh exactly the same amount. Second, newly minted coins should be weighed as those that have been in circulation for a while will weigh less due to use. Finally, the precision of the scale is a factor because finding the weight to the nearest tenth of a gram may skew the results; precision to at least the hundredths, and preferably to the thousandths, of a gram is required.

Answer and Solution They weigh the same. That’s right—trick question. According the U.S. Mint specifications, a dime weighs 2.268 grams, and a quarter weighs 5.670 grams, which means that a quarter weighs 2.5 times as much as a dime, and a quarter is worth 2.5 times as much as a dime. So, a pile of dimes and a pile of quarters having the same value will have the same weight, too. Crazy, huh?

PROBLEM 5. Days are Numbered Common Core State Standards for Mathematics: 4.MD.A.2 • 5.MD.A.1 • MP.6

In the Classroom This problem is an entry-level conversion problem using naked numbers. For young students, however, it allows practice with dividing into equal groups and considering the remainder. An important concern is how to deal with the remainder. The answer given below—4 days, 4 hours— expresses the remainder in hours. It would be mathematically correct to express the answer 1 as 4  days or approximately 4.17 days, but in day-to-day use, it’s atypical to express days with 6 fractional or decimal representations. Along the same lines, it’s important for teachers to consider how time is represented in problems that they pose to students. Students absolutely need to develop fluency with both fractions and decimals, so it can be tempting to include real-world problems that include contrived measurements. For example, a problem asking, “What is the current weight of a dog who weighed 13  5 pounds 16 a month ago, but then got sick and lost 2  5 pounds?” could elicit information about students’ 8 ability to solve fraction subtraction problems. But it gives students the impression that the weight of dogs is expressed in fractional units, which is never the case! Educators need to be mindful that problems which address specific learning targets may be detrimental over the long term to developing mathematically sophisticated students.

Answer and Solution 4 days, 4 hours. There are 24 hours in a day, and 4 × 24 = 96. Because 100 − 96 = 4, there are 4 extra hours. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

53

PROBLEM 6. Walk It Off Common Core State Standards for Mathematics: 4.MD.A.2 • 5.MD.A.1 • MP.6

In the Classroom Like problem 5, Days Are Numbered, this task provides an opportunity to discuss colloquial use of mathematics. Although technically correct, it would be atypical to express the answer for this problem as 8 1 feet since lengths in the imperial system are typically expressed as feet and inches, not 3 mixed numbers. Occasionally, halves are used, but rarely. An effective representation for converting between measurements is a double number line. This is especially true when converting between measurements within the same system, where one unit is an integer multiple of the other. Such is the case with length in the imperial system, where 12 inches = 1 foot.

Answer and Solution 8 feet, 4 inches. There are 12 inches in a foot, and 8 × 12 = 96. Because 100 − 96 = 4, there are 4 extra inches.

PROBLEM 7. Release the Hounds Common Core State Standards for Mathematics: 6.RP.A.3.C • 7.RP.A.3 • MP.1 • MP.6

In the Classroom The counterintuitive answer to this problem will bring a classroom to life. Because the difference between 99 percent − 98 percent = 1 percent, and because the situation involves 100 puppies, many people incorrectly assume that the solution is to remove just one hound. Pose the problem to the class, and then monitor students as they work. Some students will use equations, some will use guess-and-test until they find the appropriate percent, some will use 54

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

ratio reasoning and convert the percentages to fractions, and some may use other strategies. Pay careful attention to both the answers and to the strategies that they are using because you’ll want to sequence their attempted solutions during the class discussion. To facilitate conversation, allow students to suggest answers, and display the suggestions for all students to see. Then, have students vote on the answer that they think is correct. Now, have students discuss their answer with a classmate who voted for a different answer. Can they come to consensus? After two minutes, have students vote again. The results will likely change in favor of one answer. Those who disagree should be allowed to present their arguments. Many times, the process of explaining an incorrect solution will be enough for some students to realize their error. To wrap things up, allow several different correct solutions to be presented as well.

Answer and Solution 50 hounds. If 99 percent of the 100 puppies are hounds, then 99 puppies are hounds. By removing 50 hounds, there will be 50 puppies left in the shelter, with 49 of them being hounds, 49 and 50 = 98 percent.

PROBLEM 8. Digital Throwback Common Core State Standards for Mathematics: 2.NBT.B.5 • 2.NBT.B.7 • 3.NBT.A.2 • MP.3

In the Classroom Student engagement is a key component of an effective mathematics classroom (Smith 2017). An effective technique to engage students—and to establish the expectation that students should set to work as soon as they get to class—is to have a nontraditional problem waiting when they arrive. This problem is compact enough that it can be printed several times on a single sheet of paper. The paper can be cut, and one strip can be placed on each student’s desk before they arrive for class. While you’re busy taking attendance, distributing materials, or with other administrative tasks, students can start to engage in mathematical thinking.

Answer and Solution 3. If addition symbols are inserted between every pair of adjacent numbers, the sum on the left side of the equation will be 45. Consequently, it becomes necessary to concatenate adjacent numbers to form two- or even three-digit numbers. One way to do this is the following: 1 + 23 − 4 + 56 + 7 + 8 + 9 = 100 Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

55

But it’s clear that with more concatenation, fewer symbols would be needed. It may be possible to concatenate the first three digits to form 123, but beyond that, concatenating three digits will form numbers that are significantly greater than 100 and will pose problems. Consequently, the best we can hope for is to form 123 with the first three numbers, and then concatenate the rest in pairs. The question is whether it will be possible to insert symbols between those numbers to get the desired result. Indeed, it is possible, and a solution with just three operators is 123 − 45 − 67 + 89 = 100.

PROBLEM 9. Freedom of Expression Common Core State Standards for Mathematics: 3.NBT.A.2 • 4.NBT.A.1 • MP.2 • MP.6

In the Classroom This problem provides two main benefits for use in the classroom: 1. It provides an opportunity to develop computational fluency. 2. It has multiple solutions. While working on this problem, students will complete numerous addition and subtraction exercises. For most students, these exercises will be undertaken willingly because there is a higher purpose: finding a solution to the problem. Some heuristics will be useful while attempting to solve this problem—for instance, noting that the sum of the units digits of the first, third, and fifth terms must equal the sum of the units digits of the second and fourth terms. That alone won’t be a sufficient strategy and some level of guess-and-test will still be necessary. But students typically find a solution within two or three attempts, which provides enough computational practice to be worthwhile but doesn’t lead to frustration or boredom.

Answer and Solution There are many solutions. Two possibilities are— 29 − 31 + 86 − 54 + 70 = 100 67 − 80 + 45 − 23 + 91 = 100

PROBLEM 10. It Doesn’t Add Up Common Core State Standards for Mathematics: 2.NBT.B.5 • 3.NBT.A.2 • MP.1 • MP.7 56

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

In the Classroom At first look, most students find it surprising that addition symbols can be inserted to form a true equation. Many students will simply insert nine addition symbols—one in each space—and be disappointed with the result. But this provides an opportunity for students to persevere, think outside the box, and learn a new term: concatenation. To concatenate is to link things or connect them, so concatenating digits is to put them side-by-side to form larger numbers. For instance, concatenating 1 and 3 produces 13. It’s concatenation that makes a solution possible. The sum of the integers from 1 to 10 is 55—see problem 21, Gauss and Check, for a formula to compute that sum—which is 45 short of the goal. If two consecutive digits in the list, a and b, are concatenated instead of added, then the result is increased by 10a − a, because a becomes the tens digit in a two-digit number with units digit b. Once that realization is made, the solution appears: Using 5 as a tens digit increases the result by 45, and the answer is found.

Answer and Solution 10 + 9 + 8 + 7 + 6 + 54 + 3 + 2 + 1 = 100. Interestingly, there is only one solution.

PROBLEM 11. Give and Take Common Core State Standards for Mathematics: 2.NBT.B.5 • 2.NBT.B.7 • 3.NBT.A.2 • MP.2

In the Classroom This problem lends itself to using the strategy of only presenting the problem stem. That is, ask students to see what values are possible when addition and subtraction symbols are inserted into the following expression: 9 8 7 6 5 4 3 2 1 This allows all students to have an entry point as even struggling students could randomly insert addition and subtraction symbols and use a calculator to determine the result. Some students will immediately insert eight addition symbols, thereby generating a triangular number. A fair response is “That’s pretty big. But is there a way to get an even greater result?” From there, the possibilities are nearly limitless: What’s the greatest possible result? The least possible result? The least possible result that’s positive? Is it possible to get zero? Is it possible to create an expression equal to your favorite number? And then, finally, is it possible to get exactly 100? Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

57

A discussion about whether it’s possible to get zero may be enlightening for students. While it’s not possible to get zero if symbols are inserted between every pair of numbers—in that case, only odd totals are possible—it is possible if numbers are concatenated, for example, 9 + 8 + 7 + 6 + 5 − 4 − 32 + 1 = 0. Once that door is opened, students will have an easier time finding an expression equal to 100. By presenting the problem with just the stem and no question, students with math anxiety are free to explore without the pressure of having to find the answer. After several minutes of free exploration, though, most students will have an understanding of how concatenating different numbers and changing symbols between addition and subtraction affect the result, so they’ll be able to make progress toward finding an expression equal to 100.

Answer and Solution There are 10 expressions that yield a value of 100: 1. 98 − 7 − 6 + 5 + 4 + 3 + 2 + 1 = 100 2. 98 − 7 + 6 – 5 + 4 + 3 + 2 – 1 = 100 3. 98 − 7 + 6 + 5 − 4 + 3 − 2 + 1 = 100 4. 98 + 7 − 6 − 5 + 4 + 3 − 2 + 1 = 100 5. 98 + 7 − 6 + 5 − 4 − 3 + 2 + 1 = 100 6. 9 + 8 + 76 + 5 − 4 + 3 + 2 + 1 = 100 7. 9 + 8 + 76 + 5 + 4 − 3 + 2 − 1 = 100 8. 98 − 7 − 6 − 5 − 4 + 3 + 21 = 100 9. 9 − 8 + 76 − 5 + 4 + 3 + 21 = 100 10. 9 − 8 + 7 + 65 − 4 + 32 − 1 = 100 When switching from just addition as in problem 10, It Doesn’t Add Up, to both addition and subtraction, the possibilities are multiplied. Consequently, this problem has many solutions. One strategy is to insert all addition symbols, compute the result, and then modify to find a solution. Unfortunately, 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45, so changing any of the signs to subtraction will reduce a value that is already less than 100. There has to be a different approach. And indeed, there is. A solution can only be found if digits are concatenated, and that leads to a strategy. Take the simplest case and concatenate the first two numbers to form 98, then insert only addition symbols: 98 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 126 58

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

This expression is 26 more than is needed. To correct for that, find a subset of addends within the expression that have a sum of 13; by changing the symbols in front of them from addition to subtraction, the result will change by 2 × 13 = 26, and a solution will be found. For better or worse, there are multiple subsets with a sum of 13: (7, 6), (7, 5, 1), (7, 4, 2), (6, 5, 2), and (6, 4, 3). That yields five solutions: 98 − 7 − 6 + 5 + 4 + 3 + 2 + 1 = 100 98 − 7 + 6 − 5 + 4 + 3 + 2 − 1 = 100 98 − 7 + 6 + 5 − 4 + 3 − 2 + 1 = 100 98 + 7 − 6 − 5 + 4 + 3 − 2 + 1 = 100 98 + 7 − 6 + 5 − 4 − 3 + 2 + 1 = 100 Concatenating the second and third digits to form 87 and inserting only addition symbols gives the following: 9 + 87 + 6 + 5 + 4 + 3 + 2 + 1 = 117 Because the sum is odd, it won’t be possible to change symbols and get a result of 100. If the symbol in front of n is changed from addition to subtraction, the value of the expression will be decreased by 2n, an even number; and if an even number is subtracted from an odd number, the result will be odd. This is an important insight. If some digits are concatenated and only addition symbols are inserted, the sum must be even for a solution to exist. (That’s not to say that a solution will exist; but it’s impossible for a solution to exist if the sum is odd.) Further, the sum must be greater than 100, because it’s not possible to increase the value by changing symbols from addition to subtraction; and at most, two pairs of numbers can be concatenated because concatenating three pairs of numbers or concatenating three numbers to form a three-digit result will give a sum from which it is too great to recover. Consequently, the following are the only concatenations that need to be considered:

98 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 126



9 + 8 + 76 + 5 + 4 + 3 + 2 + 1 = 108



98 + 7 + 65 + 4 + 3 + 2 + 1 = 180



98 + 7 + 6 + 5 + 43 + 2 + 1 = 162



98 + 7 + 6 + 5 + 4 + 3 + 21 = 144



9 + 87 + 6 + 54 + 3 + 2 + 1 = 162



9 + 87 + 6 + 5 + 4 + 32 + 1 = 144



9 + 8 + 76 + 5 + 43 + 2 + 1 = 144



9 + 8 + 76 + 5 + 4 + 3 + 21 = 126



9 + 8 + 7 + 65 + 4 + 32 + 1 = 126



9 + 8 + 7 + 6 + 54 + 3 + 21 = 108 Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

59

Those equations in bold can be modified to yield a solution. The full list of expressions that yield 100 is given at the start of the Answer and Solution section.

PROBLEM 12. What’s It Gonna Take? Common Core State Standards for Mathematics: 2.NBT.B.5 • 2.NBT.B.7 • 3.NBT.A.2 • MP.3

In the Classroom Although this problem has a “closed end”—every student is looking for the same answer—it has an “open beginning” and an “open middle”—how students start can be different, and the strategy they use to find a solution may be different, too. More important, the solution strategy is not suggested by the presentation of the problem, so students are able to choose their own path. As with problem 11, Give and Take, this problem provides an entry point for all students. Students can explore with any set of descending integers, randomly insert addition and subtraction symbols, see what happens, and then modify as appropriate. This can be approached with a think-pair-share routine, where students first spend a few minutes exploring individually before discussing their results with a partner. Then, after a few more minutes of working in pairs, each pair can share with the entire class. It’s likely that multiple students will find expressions totaling 100 that can be made with fewer than nine integers. During the whole-class discussion, the various expressions totaling 100 can be written on the board. From this list, it will be easy to see which expression contain the fewest digits. At that point, it’s appropriate to ask two different questions: “Is it possible to make an expression with fewer digits than any of the ones listed?” and “How will you convince yourself that an expression with fewer digits is not possible?” The latter question is one with which mathematicians wrestle all the time, and students should spend some time trying to convince themselves. The solution below provides one justification.

Answer and Solution Six digits. 65 + 4 + 32 − 1 = 100. Clearly, with only five digits, things just wouldn’t work out; concatenating three digits would put the total well over 100, and the highest possible value when concatenating just two digits would be 54 + 32 + 1 = 87. It’s interesting that just one more digit allows for a solution to be found.

60

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 13. Times Square Common Core State Standards for Mathematics: 4.OA.B.4 • 4.NBT.B.5 • MP.7

In the Classroom Multiplication tables have been used for centuries as a concise representation of multiplication facts. Unfortunately, they’ve also been abused for decades in mathematics classes as a mechanism for forcing students to memorize those facts. This problem shows just one opportunity to use the multiplication table for richer tasks. The following questions provide students an opportunity to think more deeply about the relationships in the table. As a result, students will attach meaning to the values, which encourages long-term retention. n What symmetry exists in the table? Why? (The square numbers [1, 4, 9, .

. ., 100] occur along the diagonal, and there is reflection symmetry across that diagonal; that happens because multiplication is commutative, for example, 8 × 3 = 3 × 8.)

n What do you notice about all the numbers along the diagonal from 7 × 1 up to 1 × 7?

(The positive difference between the terms is 5, 3, 1, 1, 3, 5, respectively, again demonstrating symmetry, but also illuminating a number pattern.) n

What kinds of numbers occur in the row and column for 2? (doubles, or even numbers)

n

 ow many numbers end in 0? (27: There are 19 that are multiples of 10, and there are 10 that H are the product of 5 and an even number, but that counts 50 twice in both a row and column.)

To further explore patterns in the multiplication table, students can access the online Times Table from Illuminations at https://www.nctm.org/Classroom-Resources/Illuminations/Interactives/ Times-Table.

Answer and Solution 3,025. The first row is equal to 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55. The second row is 2 × 55, the third row is 3 × 55, and so on, until the tenth row is 10 × 55. In other words, the total sum is equal to (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) × 55 = 55 × 55 = 3,025.

PROBLEM 14. Squares and Square Roots Common Core State Standards for Mathematics: 8.EE.A.2 • MP.6 Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

61

In the Classroom It is perhaps an overstatement to call this a problem; for most students, it is likely an exercise. Its use in the classroom highlights important properties of squares and square roots: in particular, that squaring a positive integer will yield a result greater than the original number and that taking the square root of a positive integer will yield a result less than the original number. Squaring a number and taking the square root are inverse operations. Although young students may not yet know the concept of an inverse, it can be powerful to show that these two operations, performed in succession, will return the original number. For instance, 212 = 21 and 52  = 5.

Answer and Solution 9,990. The difference is 1002 − 100 = 10,000 − 10 = 9,990.

PROBLEM 15. What Would Your Computer Think? Common Core State Standards for Mathematics: 4.OA.C.5 • 4.NBT.A.2 • MP.7

In the Classroom Every “What do you think comes next. . .?” question is useful as a daily warm-up. This particular pattern is based on binary numbers—numbers expressed in base 2—that are used in computers and computer-based devices. But even students unfamiliar with the binary system can find the next number in the pattern if they interpret it as numbers in ascending order that can only be formed with 0s and 1s.

Answer and Solution 101. It is the sequence of binary numbers. The next number is 1012 = 510.

PROBLEM 16. Desmos Be the Place Common Core State Standards for Mathematics: 4.OA.C.5 • HSF.BF.A.1 • MP.7

In the Classroom If the only reason that this problem appears in your classroom is to provide an excuse to navigate students to the Desmos graphing calculator, then it will have been worthwhile. Interestingly, if 62

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

the terms in this sequence are plotted as (1, 2), (2, 5), (3, 10), (4, 20), (5, 50), and (6, 100), and the x curve y = b   is plotted on the same graph, using the slider reveals that b ≈ 2.154 provides a reasonable approximation to the data. That’s because the terms in the sequence increase tenfold every three 1 3 terms, and 10 ≈ 2.154. 100

50

0

2

4

6

Answer and Solution 200. Go to www.desmos.com/calculator, and you will notice that the first label on the x-axis to the right of the origin is 1. If you click the zoom out (“−”) button, the first label to the right of the origin changes to 2. Continue clicking, and it will increase to 5, 10, 20, 50, 100, 200. Two-thirds of the terms are double the preceding term, but the other third are two-and-a-half times the preceding term. The pattern is meant to keep nice numbers along the axes.

PROBLEM 17. Even I Know This One Common Core State Standards for Mathematics: 4.OA.C.5 • 8.EE.A.2 • HSF.BF.A.1 • MP.3

In the Classroom For elementary school students, this problem provides an opportunity to apply pattern recognition; the differences between consecutive terms are 12, 20, 28, 36, which increase by 8 each time. The next increase should be 36 + 8 = 44, so the next term would be 144. For middle school students, this pattern problem provides an opportunity to apply square roots; the square root of the terms is 2, 4, 6, 8, 10, a pattern that is easy to identify. For high school students, it’s an opportunity to build a function, as plotting the points (1, 4), (2, 16), (3, 36), (4, 64), (5, 100) will reveal points on the curve y = (2x)2 = 4x2. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

63

Answer and Solution 144. The terms of this sequence are the even numbers squared. The sixth even number is 2 × 6 = 12, and 122 = 144.

PROBLEM 18. Number of Numbers Common Core State Standards for Mathematics: 4.OA.C.5 • 5.OA.B.3 • HSF.BF.A.1.A • MP.8

In the Classroom Typically, students see the pattern in this problem fairly quickly, noting that there is one 1, two 2s, three 3s, and so on. What poses a more difficult challenge if they are unfamiliar with triangular numbers is determining the 100th number using brute force. Ask students to determine the position of the last 1, the last 2, the last 3, and so forth. Students will start to see a pattern emerge: 1, 3, 6, . . ., where the difference between consecutive terms is one more than the difference between the previous pair. Though not a formula, this pattern will allow students to more quickly determine that 14 occupies positions 92 through 105.

Answer and Solution 14. In this sequence, the last occurrence of digit k appears as the kth triangular number term. For instance, the last occurrence of 3 is in the sixth position, and 6 is the third triangular number. The 105th term in this sequence will be the last occurrence of 14, which means 14 will also occupy the 100th term. Students unfamiliar with the triangular numbers, though, could arrange the terms as follows to see a pattern emerge: 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5

1 3 6 10 15

The bold numbers to the right indicate the total number of terms. That is, the last 4 occupies the 10th position, the last 5 occupies the 15th position, and so on. As previously noted, the last 14 occupies the 105th position, so a 14 would occupy the 100th position, also. 64

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 19. What in the World? Common Core State Standards for Mathematics: 4.OA.B.4 • 4.OA.C.5 • MP.7

In the Classroom The pattern underlying this sequence is the prime numbers, and every student can use the sieve of Eratosthenes to determine which numbers are prime. Provide each student with a list of the integers from 2 to 100. Circle the number 2, as it is the first prime number on the list; then, cross out every multiple of 2 thereafter because we know those numbers are divisible by 2 and therefore are not prime. Now, return to the start of the list and circle the first number greater than 2 that has not been crossed out; in this case, it’s 3. Circle the number 3 as it must also be prime, because it is not a multiple of any smaller prime number. But then, cross out every multiple of 3 thereafter. Return to the start of the list, and circle the first number remaining, and then cross out every multiple of that number. Continue until every number on the list has either been circled or crossed out. The resulting list will contain only prime numbers.

Answer and Solution 25. There is no shame in your game if you didn’t decipher this pattern. The numbers in the sequence increase by 1 at terms 2, 3, 5, 7, 11, 13, . . ., which is likely a pattern that you recognize. The nth term of this sequence represents the number of prime numbers less than or equal to n. For instance, the fourth term is 2 because there are two prime numbers less than or equal to 4, namely 2 and 3; similarly, the 17th term is 7 because there are seven prime numbers less than or equal to 17, namely 2, 3, 5, 7, 11, 13, and 17.

PROBLEM 20. Farey Tales Common Core State Standards for Mathematics: 4.OA.C.5 • 4.NF.A.2 • 6.RP.A.3 • MP.7

In the Classroom Fraction clotheslines have become an increasingly popular part of the curriculum. A lesson by Heitschmidt (2008) appeared on NCTM’s Illuminations website at least a decade ago, though similar activities appeared in the Connected Math Project (1998) and other middle school curricula much earlier. More recently, Shore (2018) published an entire volume on how to use a clothesline for developing number sense with students throughout K-12. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

65

This problem lends itself nicely to using a fraction clothesline, though a rather long clothesline and many index cards or small pieces of paper will be needed. The clothesline can be pre-populated with 0 and 1, and students can place 98 index cards containing the unit fractions from 1 to 1 . 99

2

At that point, two questions can be asked. First, “How many index cards have been placed on the clothesline?” So far, 100 index cards appear on the number line. This is important for the purposes of finding an answer to the problem. Because the question asks for the 100th term of F100, the answer will clearly be less than 1 . As the activity continues, it will be necessary to periodically determine how many 2 index cards have been placed; students can be prompted to think about when they will be able to stop placing index cards because an answer to the question can be found. Second, “Are there any fractions less than 1 that are missing from the clothesline?” The answer to this question, of course, is that many 2 are missing, so a discussion of which ones are missing and how to place them can occur. Students can discuss possible strategies for this. A possible follow-up question that may provide some guidance is “What is the smallest fraction missing from the clothesline?” As this activity will only involve fractions with denominators less than 100, students should realize that 2 is the smallest missing fraction; it has 99 the largest possible denominator and the least possible numerator that hasn’t yet been used. A rich, whole-class discussion can then occur about where to place 2 on the clothesline. 99

With 2 on the clothesline, other fractions with numerator 2 and odd denominators can then 99 be placed. (Fractions with numerator 2 and even-numbered denominators are already represented on the number line by equivalent fractions whose numerator is 1.) Students can then place fractions with numerator 3 and denominators that are not multiples of 3. 3 The important point here is that once is placed, no other fraction will be placed to its left. That 89 3 is, all other fractions with denominator 3 will be placed to the right of , and the smallest fraction 89 with numerator 4, namely 4 , is greater than 3 . Other fractions may be placed on the clothesline to 89 99 extend the activity, but an answer to the original problem has been found.

Answer and Solution 3 . The Farey sequence F100 begins as follows: 89 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 1 100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 2 1 2 1 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 99 49 97 48 95 47 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 3 1 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 93 46 91 45 89 44 87 43 85 42 83 41 81 40 79 39 77 38 75 37 73 36 71 35 69 34 67 100 33 3 2 9 1 3 2 3 1 3 2 3 1 3 2 3 1 3 2 3 1 3 2 3 1 3 2 3 1 3 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 98 65 97 32 95 63 94 31 92 61 91 30 89 59 88 29 86 57 85 28 83 55 82 27 80 53 79 26 77 2 3 1 4 3 2 3 4 1 4 3 2 3 4 , , , , , , , , , , , , , , 51 76 25 99 74 49 73 97 24 95 71 47 70 93 By examining the sequence, it’s clear that the 100th term is 3 . 89

66

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 21. Gauss and Check Common Core State Standards for Mathematics: 3.NBT.A.2 • HSA.SSE.A.2 • MP.7

In the Classroom An exaggerated, if not apocryphal, story claims that Carl Friedrich Gauss, when asked by his schoolmaster to find the sum 1 + 2 + 3 + ⋯ + 100, did so in just a few seconds. Various accounts of this story place Gauss’s age at the time between 3 and 10 years; further, the stories differ in the sum that he was asked to compute, from the first 20, to the first 99, to the first 100 positive integers. What cannot be disputed, however, is that Gauss was a prodigious talent, and he likely did calculate this sum at a very early age, even if the exact details of the story have morphed over time. Moreover, this is a good story to relay to students, because it demonstrates that looking for patterns—that is, working smarter, not harder—is an effective strategy for solving difficult problems. The following account from Hawking (2005) is a concise rendition that includes both the crux of the story as well as the mathematical insight upon which Gauss relied. When the class began to be unruly, the teacher, J. G. Büttner, assigned them the task of adding up all of the integers from 1 to 100. As his classmates struggled to fit their calculations on their individual slates, Gauss wrote down the answer immediately: 5,050. As soon as the problem was stated, Gauss recognized that the set of integers from 1 to 100 was identical to 50 pairs of integers each adding up to 101. (p. 563)

Most students will be overwhelmed by the prospect of adding 100 numbers. When presenting this problem to students, acknowledge that they are welcome to add the numbers one by one, but suggest that their time might be better served by first identifying any patterns. The power of this problem in the classroom is that a rich discussion can be orchestrated about the variety of strategies used to find the sum. The role of the teacher is to elicit these strategies and assist students in seeing the connections between them. In addition to the method used by Gauss, the following are some other strategies that students may employ: n

 he sum can be rewritten as groups of five: (1 + 2 + 3 + 4 + 5) + (6 + 7 + 8 + 9 + 10) + ⋯, and T each group has a sum of 25k + 15. The consecutive sums of the groups are 15, 40, 65, . . ., each time increasing by 25.

n The sum can be rewritten as groups of 10, and the consecutive sums of the groups are 55, 155,

255, . . ., each sum having the form 100k + 55. Hence, the sum of the series is 10 × 55 + (100 + 200 + 300 + ⋯ + 900) = 550 + 4,500 = 5,050. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

67

n The average value of a term in the series is 50.5. Because there are 100 terms, the sum is

100 × 50.5 = 5,050. n

 he sum can be rewritten as (1 + 99) + (2 + 98) + (3 + 97) + ⋯ + (49 + 51) + 50 + 100, where T there are 49 pairs that add to 100. The sum of the series is 49 × 100 + 50 + 100 = 5,050.

Each of these methods, as well as others that students may find, are valid and can provide insight. Allowing students to share their methods with one another provides more mathematical tools with which to fill their toolbox.

Answer and Solution 5,050. Note that 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, . . ., 50 + 51 = 101. There are 50 pairs equal to 101, so the total sum is 50 × 101 = 5,050. 1

This method leads to the general formula 1 + 2 + 3 + ⋯ + n = (n)(n + 1), which can be proven with 2 induction. Less formally, the following representation is usually sufficient to convince students that the formula will always work:

1 n

+ 2 + 3 + ⋯ + + n − 1 + n − 2 + ⋯ +

n 1

(n + 1) + (n + 1) + (n + 1) + ⋯ + (n + 1) The sum consists of n pairs, each equal to n + 1, but each term has been counted twice, so it’s necessary to multiply by 1 . 2

PROBLEM 22. Getting Even Common Core State Standards for Mathematics: 3.OA.B.5 • 3.NBT.A.2 • HSA.SSE.A.2 • MP.7

In the Classroom This problem works well when used in tandem with problem 21, Gauss and Check. Students can apply the same strategy to find pairs with equivalent sums: 2 + 100, 4 + 98, 6 + 96, and so on, or they can first use the distributive property to realize that the expression can be rewritten as 2 × (1 + 2 + 3 + ⋯ + 50), and then use the pairing strategy or the general formula to find the sum within the parentheses.

68

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Answer and Solution 2,550. Note that 2 + 100 = 102, 4 + 98 = 102, 6 + 96 = 102, . . ., 50 + 52 = 102. There are 25 pairs equal to 102, so the total sum is 25 × 102 = 2,550.

PROBLEM 23. Well, That’s Odd Common Core State Standards for Mathematics: 3.NBT.A.2 • HSA.SSE.A.2 • MP.7

In the Classroom This problem is similar to problem 21, Gauss and Check, and problem 22, Getting Even, but it is different enough that alternative strategies can be used to solve it. It works well in the classroom because students will approach it using various methods, especially if the classroom culture has been built to value the presentation of multiple strategies. A pairing strategy yields the interesting result (1 + 199) + (3 + 197) + (5 + 195) + ⋯ + (99 + 101), which consists of 50 pairs, each with a sum of 200. The total, therefore, is 50 × 200 = 10,000. Though valid, this method fails to provide the insight that might be gained by using other methods. One rewriting strategy allows the terms of the expression to be rewritten as (1 + 0) + (1 + 2) + (1 + 4) + ⋯ + (1 + 198) = 100 × 1 + (0 + 2 + 4 + ⋯ + 198). Similarly, it could be rewritten as (2 − 1) + (4 − 1) + (6 − 1) + ⋯ + (200 − 1) = (2 + 4 + 6 + ⋯ + 200) − 100 × 1. A visual strategy allows students to see that the sum of consecutive odd numbers yields a square:

⋯

⋮

⋱

The sum 1 + 3 + 5 + ⋯ + (2 × 100 + 1) can be represented visually as a square with 100 unit squares along each side, and the total number of unit squares is 1002 = 10,000.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

69

Answer and Solution 10,000. This is the sum of the odd integers from 1 to 199. The sum of the first n odd integers is equal to n2, for instance, 1 + 3 = 22, and 1 + 3 + 5 = 32. Consequently, the given sum is equal to 1002 = 10,000.

PROBLEM 24. Square Deal Common Core State Standards for Mathematics: HSA.REI.C.6 • HSA.REI.C.8 • HSF.BF.A.1 • MP.7

In the Classroom An important concept is finite differences. A finite difference of a function is the difference between the values of the function for two different inputs. Most of the time, the difference under consideration will be the difference between the value of the function at consecutive integers. If the differences between consecutive terms of a sequence have a constant value, then the terms of that sequence can be represented by a linear function. For instance, in the sequence 2, 9, 16, 23, . . ., the difference between every pair of consecutive terms is 7. Consequently, the terms of the sequence can be described by the linear function 7n + 2. If the differences of the differences (known as the second differences) have a constant value, then the terms of the sequence can be represented by a quadratic function. More generally, if the nth differences between consecutive terms is constant, then the terms of the sequence can be represented by an nth-degree polynomial function. Ask students to generate the first several terms of the sequence from this problem, the sum of the square numbers. Then, have them find the first, second, and third differences. If students pause between each set of differences, a discussion can occur about the values in that set of differences. For this sequence, as shown below, the first differences are the square numbers, the second differences are the odd numbers, and the third differences are constant. Students are often amazed to find that one of these sets leads to a constant value. For the sequence represented in this problem—the sum of the square numbers—the third differences have a constant value.

70

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Sequence

1

First Differences Second Differences Third Differences

5 4

14 9

5

30 16

7 2

55 25

9 2

91 36

11 2

Consequently, the terms of this sequence can be represented by a third-degree polynomial. The method for determining the coefficients of that polynomial is given in the solution below.

Answer and Solution 338,350. If you know the formula for the sum, then this problem is trivial. So, the real question is how to determine a closed-form formula for the sum of the first n square numbers. The first several terms of the sequence of the sum of the square numbers are 1, 5, 14, 30, 55, 91, and so on. The sequence of first differences—that is, the differences of the terms of the original sequence—are 4, 9, 16, 25, 36, and so on—as you probably could have predicted. The second differences—that is the differences between the first differences—are 5, 7, 9, 11, and so forth, which are the odd numbers. Finally, the third differences are 2, 2, 2, and so forth. Because the third differences are a constant, then a third-degree closed-form formula exists. Now, we just need to find it. The formula for the sum S will have the form S(n) = an3 + bn2 + cn + d. This gives us the following: S(1) = a + b + c + d = 1 S(2) = 8a + 4b + 2c + d = 5 S(3) = 27a + 9b + 3c + d = 14 S(4) = 64a + 16b + 4c + d = 30 If you’ve been able to follow along to this point, then solving that system of four equations in four 1 1 1 unknowns should be possible. Doing so yields a = 3, b = , c = 6 , and d = 0. Substituting those values 2 into the general formula gives the following: n 6

S(n) =  (n + 1)(2n + 1) So the sum of the first 100 square numbers is equal to the following: S(100) =

100 100  (100 + 1)(2 × 100 + 1) = 6 × 101 × 201 = 338,350 6

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

71

PROBLEM 25. Non-Square Numbers Common Core State Standards for Mathematics: 4.NBT.B.5 • 8.EE.A.2 • MP.7

In the Classroom 10

The 10th square number is 102 = 100. Consequently, only = 10 percent of the first 100 numbers 100 100 = 1 percent are square numbers. Similarly, the 100th square number is 1002 = 10,000, so only 10,000 of the first 10,000 numbers are square numbers. Beyond that, the percentage decreases even further. As a result, students might assume that non-square numbers are somewhat rare, as well. It is therefore good to give students an opportunity to provide estimates for the 100th non-square number based on their initial thinking. Students will often reason that the 100th non-square number must be greater than 100 but less than 1002 = 10,000. Push them to improve those bounds by considering the density of the non-square numbers. Unlike the square numbers, the non-square numbers occur in big chunks of consecutive positive integers.

Answer and Solution 110. There are n2 − n non-square numbers less than n2. For example, there are 42 − 4 = 12 non-square numbers less than 42 = 16. When n = 10, n2 − n = 102 − 10 = 90. Consequently, there are 90 non-square numbers less than 100. Since there is not another square number until 121, the 100th non-square number is 110.

PROBLEM 26. In Perfect Harmony Common Core State Standards for Mathematics: 5.NF.A.1 • MP.5

In the Classroom The harmonic series is a divergent infinite series; that is, if the series continued to infinity, its value would grow without bound. This problem contains the first 100 terms of the series, and determining the sum is not an easy thing to do. But given the importance of the harmonic series— it’s a divergent series, even though the value of its terms converges to zero—an early introduction can set the stage for deeper explorations in the future.

72

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

When considering how to find the sum in this problem, it’s possible to provide a reasonable estimate. p Replace each term 1 by the fraction 1p , where p is the largest possible value such that 2  ≤ n. When n 2 these replacements are made, the original series is greater than the revised series; that is— 1 1 1 + + + 1 2 3 1 1 + + 1 2

1 1 1 + + + 4 5 6 1 1 1 + + + 4 4 8

1 1 1 1 + + + +> 7 8 9 10 1 1 1 1 1 + + + + + 8 8 8 16 16

The value of the revised series will be 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 ,, which is a little greater than 4. Since 2 2 2 2 2 2 32 this is clearly an underestimate, it would be reasonable for a student to think that the sum might be greater than 5, which is indeed the case. The solution below explains how a more accurate estimate can be obtained. This problem also provides an opportunity for engaging in MP.5 by using a spreadsheet. By placing 1 the integers 1 to 100 in column A, the formula “= ” in cell B1, and then copying that formula into A1 cells B2 through B100, the first 100 terms of the series can be generated. The sum of that sequence can be found with the formula “=sum(B1:B100)” to get a very close approximation.

Answer and Solution 5. The sum is the first 100 terms of the harmonic series. Unfortunately, calculating the sum of the first n terms of the harmonic series isn’t easy; Excel or some other spreadsheet software is recommended. But the pattern of values of n at which the sum surpasses consecutive integers is wellknown and represented in the following table: Integer k

1

2

3

4

5

6

1

4

11

31

83

227

First value of n at which the sum

1 1 1 1 + + +  + exceeds the value 1 2 3 n

of k

⋯ ⋯

More terms in this sequence can be found at https://oeis.org/A004080. That said, a reasonable (though not perfect) estimate for the sum of the first n terms of the harmonic series is given by ln(2n + 1); until n gets very large, the estimate is greater than the exact sum by approximately 0.1159. When n = 100, the estimate is ln(2 × 100 + 1) = ln(201) ≈ 5.3033, and making the correction gives 5.3033 − 0.1159 ≈ 5.1874.

PROBLEM 27. The Great Divide Common Core State Standards for Mathematics: 4.OA.B.4 • 4.NBT.B.6 • MP.7 Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

73

In the Classroom Most students will be able to solve this problem by brute force, and the majority of them will identify five factor pairs: 1 × 100, 2 × 50, 4 × 25, 5 × 20, and 10 × 10. This may lead them to think that there are 10 divisors, but 10 is a repeated factor, so there are only nine. As a follow-up to this problem, have students identify the number of factors for every integer 1 to 50. On several pieces of chart paper, make two-column tables that show each integer with its corresponding number of factors. Pre-populate the chart with the number of factors for 1 to 20. Then, assign the integers 21 to 50 to students. Ask students to determine the number of factors for each integer they are assigned and to have a partner check their work. When they’re convinced that they have correctly determined the number of factors, they can confirm with you and write their answers on the chart(s). Once all students have contributed to the chart(s), give the class two minutes of individual think time to look for patterns; then give them three minutes to share their observations with a partner. Among the observations that students may make are the following: n

Many numbers have just two factors. (prime numbers)

n

The number with the most factors is 48. (10)

n Most numbers have an even number of factors, but 1, 4, 9, 16, 25, 36, and 49 have an odd

number of factors. (square numbers) n

Only 4, 9, 25, and 49 have exactly three factors. (prime numbers squared)

Students should also find the prime factorization of each integer they were assigned and again look for patterns. Some students will notice that numbers with just one prime number in its prime factorization, such as 9 = 32 and 32 = 25, have a number of factors that is 1 more than the exponent in the prime factorization; for instance, 9 has 2 + 1 = 3 prime factors, and 32 has 5 + 1 = 6 prime factors. With some further exploration, students may notice that the number of factors in other numbers is equal to the product of 1 more than each exponent. For example, the prime factorization of 48 = 24 × 31, and 48 has (4 + 1)(1 + 1) = 5 × 2 = 10 factors; similarly, the prime factorization of 30 = 21 × 31 × 51, and 30 has (1 + 1)(1 + 1)(1 + 1) = 8 factors.

Answer and Solution Nine divisors. The divisors are 1, 2, 4, 5, 10, 20, 25, 50, and 100.

PROBLEM 28. Factor Fiction Common Core State Standards for Mathematics: 4.OA.B.4 • 4.NBT.B.6 • MP.7

74

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

In the Classroom This problem turns problem 27, The Great Divide, on its head. Instead of asking for the number of divisors in a number, it asks for a number with a certain number of divisors, which is much more difficult to solve. Students can investigate the relationship between the number of factors and the prime factorization of a number. In general, if the prime factorization is paqb, then the number of factors is (a + 1) × (b + 1). The factors of the number will take the form pmqn where 0 ≤ m ≤ a and 0 ≤ n ≤ b, and all possibilities are shown in the following chart: p0

p1

p2

q0

p0q0

p1q0

p2q0

q1

p0q1

p1q1

p2q1

q2

p0q2

p1q2

p2q2

⋮

⋮

⋮

⋮

qb

p0qb

p1qb

⋯

pa

⋯

paq0

⋱

⋮

⋯

paq1

⋯

paqb

⋯

p2qb

paq2

As a more concrete example, consider the number 24 = 23 × 31. It has (3 + 1) × (1 + 1) = 8 factors, as shown below: 20

21

22

23

30

1

2

4

8

31

3

6

12

24

This idea can be generalized for numbers with more than two prime factors.

Answer and Solution 43,560. A number with exactly 100 factors has the form paqbrcsd where (a + 1) × (b + 1) × (c + 1) × (d + 1) = 100. The smallest possible number is 71513424 = 43,560. Interestingly, 43,560 is also the number of square feet in an acre. The fact that it has 100 factors is likely part of the reason that it was chosen as the standard size. If the nicer number 40,000 = 2654 had been used instead, there would have been fewer possible integer divisions since 40,000 has only 35 factors.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

75

PROBLEM 29. The Locker Problem Common Core State Standards for Mathematics: 4.OA.B.4 • MP.4

In the Classroom If ever a problem were born for the act-it-out strategy, this is it. One possibility is to number index cards or small pieces of paper from 1 to 100 and allow students to run a simulation at their desks. (Obviously, a smaller number of cards could be used; even with cards from 1 to 20, a pattern can be identified.) But for some real fun, invite some students to the front of the room to serve as lockers. Approximately half the class can participate as lockers. Provide each student with a number on a piece of paper, being sure they are arranged in ascending order. Then, have the other half of the class play the role of students who open and close the lockers. To demonstrate, pretend that you are the first student, and “open” every locker. Then invite another student to be the second student and “close” every second locker. And so on. The pattern of which lockers will remain open will soon emerge. Although a fun and engaging activity for students, the purpose of the problem is neither to identify the pattern nor determine an answer. Rather, students should be pressed to explain why the pattern occurs. If necessary, encourage students to investigate individual lockers to determine how many times they were opened and closed. (While acting out the problem, you can ask students serving as lockers to keep track of how many times they were opened and closed.) Students should make the connection that each locker was opened or closed one time for each factor of the number, and only square numbers have an odd number of factors.

Answer and Solution Ten lockers will be open. Consider a particular locker, say locker 12. It will be opened by student 1, closed by student 2, opened by student 3, closed by student 4, opened by student 6, and finally closed by student 12. Notice that its state was changed six times, once for each factor of 12. In general, the number of times the state of a locker will change is equal to the number of positive integer factors of that number. Consequently, the problem reduces to which numbers from 1 to 100 have an odd number of factors. Those are the only lockers that will be open after the process concludes. As it turns out, only square numbers have an odd number of factors because they have a “repeated factor” equal to the square root of the number (e.g., the square number 36 has a repeated factor of 6 because 6 × 6 = 36). When all 100 students have finished opening and closing the lockers, lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 will be open; the other 90 lockers will be closed. 76

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

The following questions can be used as extensions or for class discussion: n

Which lockers will have its state changed exactly twice? [primes]

n

Which students changed the state of both lockers 36 and 48? [1, 2, 3, 4, 6, 12]

n

Which locker will have its state changed the most times? [96]

PROBLEM 30. Zero Hero Common Core State Standards for Mathematics: 4.NBT.A.2 • 8.EE.A.1 • 8.EE.A.3 • MP.1

In the Classroom Problems about trailing zeroes are ubiquitous in math competitions, but they are also useful in standard classrooms because of what they reveal about the number system. In the early grades, students are taught the erroneous rule that one should simply “add a zero” when multiplying by 10. To see that this is erroneous, consider the product of 2.38 × 10. The rule only applies when multiplying integers. Moreover, multiplying an integer by 10k adds k trailing zeroes to the end of the number. Exploring problems like “How many trailing zeroes does 100! have?” allows students to determine the underlying structure, namely that the number of trailing zeroes is equal to min(a, b) if 2a and 5b are contained in the prime factorization.

Answer and Solution There are 200 zeroes. There is a pattern: 1002 = 10,000, which has four 0s; 1003 = 1,000,000, which has six 0s; 1004 = 100,000,000, which has eight 0s; and so on, such that 100n has 2n zeroes at the end.

PROBLEM 31. Famous Last Digits Common Core State Standards for Mathematics: 4. OA.C.5 • 8.EE.A.1 • MP.7 • MP.8

In the Classroom Every classroom should have a piece of chart paper with the title “Problem Solving Strategies We Know and Love.” As the school year progresses and more and more strategies are uncovered, they can be added to the chart. The benefit of such a chart is that it serves as a reminder of all the tools Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

77

available to students, and it helps to dispel the notion that there’s “one right way” to solve a given problem. When students are stuck, they can use the chart as a trigger to get unstuck. The strategy that can be added to the chart as a result of this problem is “look for a pattern.” As shown in the solution below, the search for a pattern bears fruit by revealing that the last two digits of powers of 2 occur in a cyclical pattern.

Answer and Solution 76. Consider the following pattern: Power of 2

Last 2 Digits

2

1

2

22

4

23

8

2

4

16

25

32

6

2

64

27

28

28

56

9

2

12

210

24

2

11

48

212

96

213

92

2

14

84

215

68

2

16

36

217

72

218

44

2

19

88

220

76

2

21

52

222

04

223

08

2

16

24

78

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Note that 24 and 224 both have 1 and 6 as the last two digits. This implies that the cycle of the last two digits repeats every 20 powers of 2. Consequently, 220, 240, 260, 280, and 2100 will all have the same last two digits, which are 7 and 6.

PROBLEM 32. More Famous Last Digits Common Core State Standards for Mathematics: 8.EE.A.1 • MP.7 • MP.8

In the Classroom The length of the equation with eight terms and three-digit exponents will be exciting for some students and overwhelming for others. Students who are excited can be encouraged to solve the problem on their own or with a partner, whereas students who need support can be involved in a group problem-solving task. The terms of the equation could be distributed, one per student, to find the units digit of the 100th power for just that number. That is, one student could find the units digit of 3100 while another student finds the units digit of 8100. The combined work of nine students will eventually lead to a solution, reinforcing that there is strength in numbers.

Answer and Solution 3. The units digits in the powers of integers always repeats in cycles of one, two, or four numbers, as shown in the table below: Exponent (n)

Units Digit of . . . 1n

2n

3n

4n

5n

6n

7n

8n

9n

1

1

2

3

4

5

6

7

8

9

2

1

4

9

6

5

6

9

4

1

3

1

8

7

4

5

6

3

2

9

4

1

6

1

6

5

6

1

6

1

5

1

2

3

4

5

6

7

8

9

6

1

4

9

6

5

6

9

4

1

7

1

8

7

4

5

6

3

2

9

8

1

6

1

6

5

6

1

6

1

Consequently, the fourth power of every integer will have the same units digit as the 100th power, and the sum is 1 + 6 + 1 + 6 + 5 + 6 + 1 + 6 + 1 = 33. Therefore, the units digit of the sum will be 3.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

79

PROBLEM 33. At the End, There’s Nothing Common Core State Standards for Mathematics: 4.OA.C.5 • 5.NF.A.1 • HSS.CP.B.9 • MP.8

In the Classroom Most technology will not calculate factorials as large as 100!, and the majority of calculators will start to report factorials larger than 15! in scientific notation, which is not helpful for determining the number of trailing zeroes. Consequently, students can look for a pattern by making an organized list. As shown below, the number of trailing zeroes in n! increases in a predictable way as n increases. Factorial

1!

2!

3!

4!

5!

6!

7!

8!

9!

10!

Number of Zeroes at the End

0

0

0

0

1

1

1

1

1

2

⋯ ⋯

The list could be continued to 100!, but eventually the calculations will need to be performed by hand, which is both inefficient and difficult. From the list, though, it seems clear that the number of trailing zeroes increases by 1 each time n reaches a multiple of 5. Students should be encouraged to consider why that happens, and brief partner discussions will generate conjectures. One likely conjecture is that the number of trailing zeroes increases each time a new factor of 5 occurs.

The important question, though, is whether this will continue to happen indefinitely. Indeed, it will with the exception that the number of trailing zeroes increases by 2 at n = 25, 50, 75, and 100. Per the conjecture above, the number of trailing zeroes will increase by 2 each time n reaches a multiple of 25, by 3 each time n reaches a multiple of 125, and so on, with the number of zeroes increasing in general by k each time n reaches a multiple of 5k.

Answer and Solution 24. The number of 0s at the end depends on how many 5s and 2s are in the prime factorization; each pair of 2 × 5 will yield a 0 at the end of the number. There are far more 2s than 5s, so we merely need to determine the number of 5s in the prime factorization to solve the problem. All 20 multiples of 5 have one factor of 5; the 4 multiples of 25 have a second factor of 5. Consequently, the number 100! will end in 20 + 4 = 24 zeroes.

PROBLEM 34. Can You Digit? Common Core State Standards for Mathematics: 8.F.A.1 • HSF.IF.A.2 • MP.2 80

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

In the Classroom The primary motivation for using this problem is that it provides an example of function notation that is approachable by students. Understanding how the function works—it simply counts the number of digits—is straightforward, so students can instead attend to the nested function notation and what that means. To extend students’ thinking, have them investigate the more general function S(S(x)) = k, for small integer values of k. For instance, what values of x are possible when k = {1, 2, 3, 4}?

Answer and Solution 3. This one is pretty straightforward if the notation doesn’t get in your way. When expanded, the number 100100 will be a 1 followed by 100 pairs of two zeroes, so it has 201 digits. Hence, S(100100) = 201. Therefore, S(S(100100)) = S(201) = 3.

PROBLEM 35. Sum Kinda Wonderful Common Core State Standards for Mathematics: 4.NBT.A.1 • 4.NBT.A.2 • MP.3

In the Classroom Though perhaps trivial for older students, this problem is great for young students to ponder after they’ve developed an understanding of place value. It’s also a great opportunity to invoke a fundamental problem-solving strategy: “solve a simpler problem.” Students can select a number to fill in the blank when given the prompt “The sum of the digits of a number is __, and none of the digits are 0. What numbers are possible?” Doing so allows for natural differentiation, though some guidance will be required to ensure that students don’t choose a number that is too large. To start, it may be necessary to limit the choices to single-digit numbers, and even then, there are 512 possible numbers if they select 9 for the blank. Once students start to list the possible numbers, a class discussion could be used to look for patterns. n There are two numbers with a digital sum of 2 and only nonzero digits, namely 2 and 11; of

these, 11 is larger. n There are four numbers with a digital sum of 3 and only nonzero digits, namely 3, 12, 21, and

111; of these, 111 is the largest. n There are eight numbers with a digital sum of 4 and only nonzero digits, namely 4, 13, 22, 31,

112, 121, 211, and 1,111; of these, 1,111 is the largest. n In general, there are 2

n

numbers with a digital sum of n and only nonzero digits; of these,

111⋯111 is the largest. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

81

Students can compare their answers with other students in the class who used the same number in the blank. Doing so allows for collaboration and makes it less likely that possible numbers will be missed. For identifying the greatest number, students will come to realize that it’s always better to have more digits, because the greater the place value of the first digit, the larger the number. Consequently, the greatest number will result when the digit 1 is repeated.

Answer and Solution 1,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111, 111,111,111,111,111,111,111,111,111,111,111,111,111. The greatest number is going to be the one with the most digits, so the correct strategy is to use one hundred 1s.

PROBLEM 36. Product Marketing Common Core State Standards for Mathematics: 4.NBT.A.2 • 4.OA.B.4 • MP.3

In the Classroom In contrast to problem 35, Sum Kinda Wonderful, this problem asks for a number with a digital product of 100, and the result is a number with many fewer digits. The solution relies on identifying the prime factorization, and then composing a number consisting only of those single-digit primes. As an extension, students can explore what other digital product values are possible. That is, it’s possible to construct a number with a digital product of 100, but is it possible to construct numbers with digital products of 18, 39, 55, or 98? Why or why not? What other digital products less than 100 are not possible?

Answer and Solution 5,522. The prime factorization of 100 is 22 × 52, so the number can be created from the four digits 2, 2, 5, and 5. The largest possible number is 5,522.

PROBLEM 37. Whole Lotta Nothin’ Common Core Standards for Mathematics: 2.NBT.A.2 • 4.NBT.A.1 • MP.8

82

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

In the Classroom On www.estimation180.com, Andrew Stadel promotes the idea of asking students for an estimate that is too high, an estimate that is too low, and an estimate that is just right—all before students attempt to solve a problem. Not only does this tactic allow students to improve their estimation skills, it also allows them to determine whether their answer is reasonable when they finally solve the problem. That routine will work well with this problem. Require students to supply estimates before putting pen to paper, and don’t accept estimates that are offered without thought. Students should be able to reason that a lower bound must be greater than 100, because the vast majority of the first 100 positive integers do not contain a 0; moreover, only some numbers in every group of 100 integers contain a 0, so a better lower bound might be 300 or higher. On the other hand, at least 10 percent of all integers in every group of 100 contain a 0, so 1,000 is a reasonable upper bound. In legitimately trying to find an estimate that is too low, but only a little bit too low, and an estimate that is too high, but only a little bit too high, students have actually already begun the problemsolving process. In attempting to find reasonable estimates, they’ve already started thinking about how many positive integers might contain a 0, and that’s a very good start.

Answer and Solution 550. From 1 to 100, there are 10 numbers that contain a 0: 10, 20, 30, . . ., 100. From 101 to 200, there are 19 numbers that contain a 0: 101, 102, 103, . . ., 110 and 120, 130, 140, . . ., 200. Likewise, there are 19 numbers from 201 to 300, 19 numbers from 301 to 400, and 19 numbers from 401 to 500. So far, that’s 10 + 19 + 19 + 19 + 19 = 86 numbers that contain a 0. There are 10 numbers with a 0 from 501 to 510, bringing the total to 96. The 97th number with a 0 is 520, the 98th is 530, the 99th is 540, and the 100th is 550.

PROBLEM 38. No Zeroes Common Core State Standards for Mathematics: 2.NBT.A.2 • 4.NBT.A.1 • MP.8

In the Classroom As with problem 37, Whole Lotta Nothin’, it would be good to give students an opportunity to suggest an estimate before they solve the problem. To extend the idea, though, categorize their estimates into ranges. For instance, the answer to this problem is 121, so the following tally chart could be used:

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

83

Estimate

Number of Students

< 50 51−100 101−200 201−300 301−500 501−1,000

More rows could be added, as necessary. The benefit of this approach occurs when the final answer is revealed. When students’ specific estimates are listed, many students feel that they estimated poorly because their estimate and the final answer disagree. In contrast, by listing estimates within ranges, students whose estimate falls within the correct range believe that their estimate was good, and students whose estimate lies in a different range will see that others estimated similarly, so it doesn’t feel like a personal failure.

Answer and Solution 121. As noted in the solution to problem 37, Whole Lotta Nothin’, there are 10 numbers from 1 to 100 that contain a 0, which means there are 90 that do not. The numbers 101 to 110 all contain a 0, but none of the nine numbers from 111 to 119 contain a 0. That brings the total to 99, and because 120 contains a 0, the 100th number without a 0 will be 121.

PROBLEM 39. Non-Zero Product Common Core State Standards for Mathematics: 4.OA.B.4 • 4.NBT.B.5 • MP.6

In the Classroom This problem is appropriate for students in intermediate classrooms who are learning or practicing multiplication. The goal for all students is to develop automaticity, which Baker and Cuevas (2018) describe as the ability to instantly recall or process information without interference from low-level details (p. 16). While it’s impossible to search the internet for “multiplication problems” and not find a worksheet of repetitive multiplication exercises, students don’t (and shouldn’t) learn their multiplication facts by memorization. Instead, interesting puzzles can be used to engage students 84

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

in practice that doesn’t feel like practice. One such puzzle is Picasso’s Line Puzzle, developed by Gordon Hamilton and described at http://mathpickle.com/project/picassos-line-puzzles. The goal of Picasso’s Line Puzzle is to completely cover a square grid numbered in a spiral pattern. The grid can be covered by paths that start at a number m, meander through n squares, and end at the square with number m × n. That is, the product of the start number and the number of squares in the path must equal the end number. For instance, in the 10 × 10 grid below, the gray path starts at 2, covers 15 squares, and ends at 30. This is acceptable, because 2 × 15 = 30. 82

83

84

85

86

87

88

89

90

91

81

50

51

52

53

54

55

56

57

92

80

49

26

27

28

29

30

31

58

93

79

48

25

10

11

12

13

32

59

94

78

47

24

9

2

3

14

33

60

95

77

46

23

8

1

4

15

34

61

96

76

45

22

7

6

5

16

35

62

97

75

44

21

20

19

18

17

36

63

98

74

43

42

41

40

39

38

37

64

99

73

72

71

70

69

68

67

66

65

100

However, paths are not allowed to reconnect, even at a corner. The black path above connects 1 to 11 with 11 squares, and 1 × 11 = 11, but the path touches corners at 9 and 11. The black path is not acceptable. The goal of the puzzle is to cover the entire grid with paths. When solving, it helps to mark each path with a different color to keep them separate. The versatility of this puzzle is that any size grid, numbered in any possible way, could be used. As a result, differentiation is naturally built into the task as students can select the grid size and how it’s numbered.

Answer and Solution 4 and 25. There are five pairs of positive integers that have a product of 100: 1 × 100 2 × 50 4 × 25 5 × 20 Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

85

10 × 10 Only 4 × 25 doesn’t have the digit 0 in either factor.

PROBLEM 40. To Say the Least Common Core State Standards for Mathematics: 2.NBT.A.1 • 2.NBT.A.2 • MP.7, MP.8

In the Classroom For most students, determining which digit appears the least number of times will not pose a great challenge. Similarly, counting the number of zeroes will not be terribly difficult, either, but doing so allows insight into the structure of our number system. In addition, it reveals patterns that students may not notice without explicit exploration. From 1 to 100, the digit 0 appears roughly half as often as the other digits. What’s surprising and somewhat counterintuitive, however, is that in a randomly generated set of data, the digits will appear in roughly equal numbers. Students are often amazed by this fact, so it’s worth investigating this claim. To do so, have students investigate the number of 0s in two-, three-, and four-digit integers. The results can be tabulated by hand, though a spreadsheet program may be helpful, especially if students would like to investigate numbers with five or more digits. n

 or two-digit numbers, 0 appears 9 times, every other digit appears 18 times. The ratio is 1:2, F or 50 percent.

n

 or three-digit numbers, 0 appears 180 times, every other digit appears 280 times. The ratio is F 9:14, or about 64 percent.

n

 or four-digit numbers, 0 appears 2,700 times, every other digit appears 3,700 times. The ratio F is 27:37, or about 73 percent.

n

 or five-digit numbers, 0 appears 36,000 times, every other digit appears 46,000 times. F The ratio is 18:23, or about 78 percent.

n

 or six-digit numbers, 0 appears 450,000 times, every other digit appears 550,000 times. F The ratio is 9:11, or about 82 percent.

As the number of digits continue to increase, the ratio of 0s to other digits will get closer and closer to 1:1. This counterintuitive result is exactly the type of thing that happens when infinity is considered. The corollary question is “What is the probability that a randomly chosen number contains the digit 0?” Choosing a number truly at random means that it could have any number of digits, and the 86

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

possibility that it would be too large to fit on a page or even in a book is very real. When viewed in that light, it doesn’t seem so unrealistic that the number might contain a 0. Given the results above, the counterintuitive answer to this corollary question is 100 percent.

Answer and Solution The digit 0 appears 11 times. From 1 to 99, each digit 1 to 9 appears 10 times as a units digit and 10 times as a tens digit. In contrast, the digit 0 appears nine times as a units digit. Even accounting for the two 0s that appear in 100, the digit 0 still appears the fewest times. Consequently, 0 appears the least number of times, just over half as many times as the other digits.

PROBLEM 41. I Got Chills, They’re Multiplying Common Core State Standards for Mathematics: 4.NBT.A.2 • 5.NBT.A.3.A • 5.NBT.B.5 • MP.7 • MP.8

In the Classroom Students will have fun exploring what happens when numbers with fewer than one hundred 9s are multiplied by 9. (See the solution below for the pattern that results.) Exploring and explaining why the products have that pattern, however, is a valuable exercise. Consider the product of 99 × 9, which can be represented as follows: 99 × 9 = 90 × 9 + 9 × 9 = 810 + 81 = 800 + (10 + 80) + 1 = 891 The expression in the parentheses yields a 9 in the tens place. Consider the product of 999 × 9, which can be represented as follows: 999 × 9 = 900 × 9 + 90 × 9 + 9 × 9 = 8,100 + 810 + 81 = 8,000 + (100 + 800) + (10 + 80) + 1 = 8,991 The expressions within parentheses contribute a 9 to the hundreds and tens places, respectively. The result can then be generalized.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

87

Answer and Solution 99. Consider the following pattern:

9 × 9 = 81



9 × 99 = 891



9 × 999 = 8,991

9 × 9,999 = 89,991 9 × 99,999 = 899,991 When a number with n 9s is multiplied by 9, the result is n − 1 9s between an 8 and a 1. Consequently, when a number with one hundred 9s is multiplied by 9, there will be ninety-nine 9s in the middle of the number.

PROBLEM 42. Ring, Rang, Rungs Common Core State Standards for Mathematics: 4.OA.C.5 • MP.8

In the Classroom Many classroom floors are covered in tiles, and the shared sides between two tiles can represent rungs of the ladder. If your floor doesn’t have tiles, use masking tape to create one or more ladders on the floor. Students can then act out the situation for small numbers in an attempt to look for a pattern. Some students will realize that the number of ways to get to each rung is the sum of the number of ways to get to the previous two rungs; other students will recognize the pattern immediately as the Fibonacci sequence. More important than seeing the pattern, though, is justifying why it occurs: It’s possible to get to any rung from the previous rung by a step size of one or from two rungs below with a step size of two. As an extension, students can do further research on the Fibonacci sequence.

Answer and Solution 354,224,848,179,261,915,075 ways. This question transmutes to determining how many combinations of 1s and 2s have a sum of 100. There are obviously a lot, since there could be fifty 2s; there could be forty-nine 2s and two 1s, which could be arranged in 51C2 = 650 different ways; there could be forty-eight 2s and four 1s, which could be arranged in 52C4 = 1,326 ways; and so on. Counting all of those possibilities could become tedious, so perhaps there are other strategies. 88

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

An alternative is to think about the number of ways to get to a particular rung. Let’s call climbing up two runs a jump, and let’s call climbing up one rung a hop. It’s possible to get to the nth rung by taking a jump from the (n − 2) rung or by taking a hop from the (n − 1) rung. Consequently, the number of ways to get to the nth step is equal to the number of ways to get to the (n − 2) step plus the number of ways to get to the (n − 1) step. There is only one way to get to the first rung. There are two ways to get to the second rung, by taking either a jump or by two hops. So, there are 1 + 2 = 3 ways to get to the third rung, by taking a hop from the second rung or by taking a jump from the first rung; there are 2 + 3 = 5 ways to get to the fourth rung, by taking a hop from the third rung or by taking a jump from the second rung; there are 3 + 5 = 8 ways to get to the fifth rung, by taking a hop from the fourth rung or by taking a jump from the third rung; and so forth. By this point, you may recognize the pattern of numbers 1, 2, 3, 5, 8, and so forth, as the Fibonacci sequence, so the number of ways to get to the 100th step is the 100th Fibonacci number. Unfortunately, determining the 100th Fibonacci number is not an easy thing to do; thanks to the list at https://tinyurl.com/ First300FibNumbers, we know that its value is 354,224,848,179,261,915,075. If you were able to climb one rung per second, if you were able to return to the bottom of the stairs instantaneously, and if you attempted to climb the stairs in every possible combination . . . it would take you a very long time, indeed!

PROBLEM 43. Leo’s Bank Account Common Core State Standards for Mathematics: 4.NBT.B.4 • 7.EE.B.4 • MP.5 • MP.8

In the Classroom This problem lends itself to three different strategies, and the approach used in the classroom should depend on the learning goals. For middle school students who are working with expressions, it’s appropriate for them to consider how the amount of money for each day of the week could be represented in terms of the amount of the deposits on Monday and Tuesday. As shown in the solution below, that’s only half the battle; once the equation 5m + 8t = 100 is found, there is still work to be done. With one equation and two unknowns, logic must be applied to find all possible answers. Alternatively, students could use a modified trial-and-check strategy to find a solution. By choosing two reasonable numbers for the Monday and Tuesday deposits, they could see what happens and then adjust. For instance, with $5 and $10 as the first two amounts, the daily deposits would be 5, 10, 15, 25, 40, 65, 105, which yields more than $100 on Sunday. Adjusting the values will eventually lead to a solution. For greater efficiency, a spreadsheet could be used so that the results from different seed values could be generated more quickly.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

89

Students might also notice the similarity between this problem and the Fibonacci sequence, leading to a third strategy. And therein lies the beauty of this problem. Students could very well use any of these strategies, but the power lies in a discussion that highlights the connections between all three.

Answer and Solution The deposits on consecutive days could have been either 4, 10, 14, 24, 38, 62, 100 dollars or 12, 5, 17, 22, 39, 61, 100 dollars. One approach is to use algebra. If Monday’s deposit was m dollars and Tuesday’s deposit was t dollars, then the following table shows the deposit each day of the week: Mon

Tue

Wed

Thu

Fri

Sat

Sun

m

t

m+t

m + 2t

2m + 3t

3m + 5t

5m + 8t

Since his deposit on Sunday was $100, we can solve 5m + 8t = 100 to find the answer. Well, sort of. You can’t solve it in the traditional sense because that’s a Diophantine equation. Instead, we have to find ordered pairs (m, t) that will satisfy the equation, and there may be more than one. In fact, a little guess-and-check reveals that there are two: 4, 10, 14, 24, 38, 62, 100 and 12, 5, 17, 22, 39, 61, 100. Alternatively, you might recognize that the pattern of Leo’s deposits mimics the Fibonacci sequence, in which each term is the sum of the previous two terms. The ratio of consecutive numbers in the Fibonacci sequence is approximately equal to the golden ratio, ϕ, which is roughly 1.618. Using that piece of information allows a working backward strategy to be employed, suggesting that the deposit on Saturday must have been approximately 100 ÷ 1.618 ≈ 61.8. As shown above, the two solutions have either $61 or $62 as the next-to-last deposit. Successive divisions by 1.618 will then reveal the entire sequence.

PROBLEM 44. Prime Time Common Core State Standards for Mathematics: 4.OA.B.4 • MP.6

In the Classroom This problem accomplishes two goals. First, it allows students to investigate numbers from 1 to 100 to determine if they are prime, which is a requirement of the Common Core State Standards for Mathematics. Second, it includes embedded skill practice with addition. Students may find prime numbers less than 50, and then determine whether the addend needed to make 100 is also prime. For instance, students might determine that 43 is prime; using an adding-on 90

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

strategy, students might then realize that 43 + 7 = 50, so 43 + 57 = 100. Now, they simply have to determine whether 57 is prime to see if they have found a solution. Alternatively, students might realize that 39 = 3 × 13, so it’s not prime. Since 39 + 61 = 100, there would be no need for them to check whether 61 is prime or not because it won’t be part of a prime pair that adds to 100.

Answer and Solution Six prime pairs. Namely, (3, 97), (11, 89), (17, 83), (29, 71), (41, 59), and (47, 53).

PROBLEM 45. In Their Prime Common Core State Standards for Mathematics: 4.OA.B.4 • MP.6

In the Classroom Very similar to problem 44, Prime Time, this problem’s key difference is that three addends are required. At first, the problem may feel overwhelming because there are 25 primes less than 100, so there are 25C3 = 2,300 combinations of three prime numbers to consider. Students might quickly jump to the conclusion that there are no solutions, incorrectly asserting that “all prime numbers are odd,” and three odd numbers will have an odd sum. But a moment of reflection should make them realize that there is one even prime number—2—and for three prime numbers to have a sum of 100, one of the addends must be even. That realization will lead students down the path to a full solution.

Answer and Solution Three sets of three prime numbers. Any set of three prime numbers with a sum of 100 must have 2 as one of the elements in the set. Consequently, this problem reduces to “How many pairs of prime numbers have a sum of 98?” Surprisingly, there are only three pairs that satisfy that criteria: (19, 79), (31, 67), and (37, 61); therefore, the three sets of three prime numbers are (2, 19, 79), (2, 31, 67) and (2, 37, 61).

PROBLEM 46. Prime Pair Common Core State Standards for Mathematics: 4.OA.B.4 • 5.NBT.B.5 • MP.6

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

91

In the Classroom This problem is a modified version of a question that originally appeared in the MathCounts School Handbook 1995−96 (p. 67), which read “Among all possible pairs of prime numbers with the sum 100, what is the largest product?” The modified wording used here serves two purposes: First, it breaks a long one-sentence problem statement into two shorter sentences, which makes it easier for students to separate the information from the question. Second, the first sentence can be used as a problem stem, and students can find pairs of prime numbers that sum to 100 (see problem 44, Prime Time) without worrying about finding the answer. It’s important to remember that changing the presentation of numbers within a problem are both rights and duties of educators to ensure that they are addressing the needs of their students. Problems don’t have to be used exactly as presented in a textbook; in fact, the problem doesn’t have to be used as presented here, either, if an alternate presentation would be more appropriate. This problem may be my all-time favorite because it primes the pump (with pun fully intended!) for some very interesting math. Beyond the initial problem, consider the following progression: n Two positive integers have a sum of 100. What’s the maximum possible product? (This is

problem 47, Max Product.) n

Three positive integers have a sum of 100. What’s the maximum possible product?

n

Four positive integers have a sum of 100. What’s the maximum possible product?

n Some positive integers add up to 100. What’s the maximum possible product? (This is

problem 48, To the Max.) n

Some positive numbers add up to 100. What’s the maximum possible product?

The problems that involve a specific number of addends have geometric analogs; the maximum product of two numbers with a given sum is equivalent to the maximum area of a rectangle with a fixed perimeter; the maximum product of three numbers is equivalent to the maximum volume of a cube. The problem becomes more difficult but more interesting when the number of integers is omitted (“some”). It becomes more interesting still when the requirement for integers is removed (“positive numbers”); it opens the door for a peek into ideas from calculus and infinity.

Answer and Solution 2,491. There are several pairs of primes that have a sum of 100: (3, 97), (11, 89), (17, 83), (29, 71), (41, 59), and (47, 53). Of those, the last has the greatest product: 47 × 53 = 2,491.

92

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 47. Max Product Common Core State Standards for Mathematics: 3.NBT.A.2 • 5.NBT.B.5 • MP.6

In the Classroom To present this task in the classroom, ask every student to think of two numbers that add up to 100. (Because many students will naturally pick 50 and 50, you could also ask them to pick a number less than 50, and then find the difference between their number and 100. That will provide a wider distribution of results to explore.) Once all students have done that, ask them to find the product of those two numbers. Then, ask them to turn and share their result with a classmate. What do they notice? What do they wonder? A good way to proceed is to ask, “What question do you think I’ll ask next?” Most students will be one step ahead of you. When they see the differences in their products, they’ll anticipate that you’ll ask them to find the greatest possible product, and that’s when the real fun begins. After students have solved the original problem, you could ask them to choose a number at random, and then solve the problem with 100 replaced by their number. When students then compare their maximum products, they’ll see that the maximum value occurs when the two factors are equal. A week or so later, it’s fun to present students with an analogous geometry problem: “The sum of the length and width of a rectangle is 100 centimeters. What is the maximum possible area of the rectangle?” During a follow-up class discussion, guide students to see that the two problems are, in fact, the same.

Answer and Solution 2,500. The maximum product will occur when the numbers are as close to one another as possible. In this case, they can be equal, and 50 × 50 = 2,500. Of note, this problem is analogous to the question, “What is the maximum area of a rectangle whose perimeter is 200 units?” The answer is 2,500 square units because the maximum area occurs when the shape is a square.

Problem 48. To the Max Common Core State Standards for Mathematics: 3.NBT.A.2 • 5.NBT.B.5 • 8.EE.A.1 • MP.6

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

93

In the Classroom This problem is another example of one that lends itself to being used as problem stem. Instead of presenting the full problem, simply ask students to find some numbers that add up to 100. This allows an entry point for all students: Even struggling students will be able to suggest that 50 + 50 can be used, and high achievers will likely opt for a more complex combination, such as 10 + 20 + 30 + 40. After all students identify at least one combination, then ask them to determine the product of their numbers. Students should absolutely be allowed to use a calculator for this investigation. The intent is for them to notice patterns and structure, not to get bogged down with calculations. As the activity progresses, the products will become unwieldy. Once students have found the product of their numbers, ask them to turn to a classmate and compare their results. There is often an eruption, and you will undoubtedly hear a student exclaim, “How did you get such a large product?” Allow for a few minutes of discussion, and then ask the real question, “What question do you think I’m going to ask next?” Most students will already know what’s coming, and they will correctly predict that the task is to see how great a product they can obtain. Then it’s off to the races as students work in pairs or small groups to see how big a product is possible.

Answer and Solution 7,412,080,755,407,364. To get a sense of what’s happening, assume that all numbers in the set have the same value. The results are (approximately) as follows: Number

How Many Needed for a Sum of 100

Product of Those Numbers

1

100

1100 = 1

2

50

250 = 1,125,899,906,842,624

3

33

333 = 5,559,060,566,555,523

4

25

425 = 1,125,899,906,842,624

5

20

520 = 95,367,431,640,625

6

17

617 = 16,926,659,444,736

7

14

714 = 678,223,072,849

It’s clear that the greatest product results when the number 3 is used. However, 100 ÷ 3 = 33 R 1, and 333 × 1 = 5,559,060,566,555,523 does not result in the largest value. It’s possible to do just a little better by using one fewer 3, and then using two 2s to get to 100. This gives the product 94

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

332 × 22 = 7,412,080,755,407,364. (It makes sense to use 3s as often as possible but not for a sum of 4, since 22 > 31, but 32 > 23.)

PROBLEM 49. Why, Certainly Common Core State Standards for Mathematics: 4.OA.A.3 • HSA.APR.A.1 • HSA.APR.C.4 • MP.3 • MP.8

In the Classroom Ask students to think of a random number. Then, have them multiply that number by the next positive integer, and then by the next, and by the next, and so on, until they get a result that is divisible by 100. (A brief discussion about how they know a number is divisible by 100 could ensue.) For some students, they may reach such a number in six steps; for others, it will take as many as nine. This allows for a discussion about how many numbers are necessary to guarantee that the product is divisible by 100, and it provides an opportunity for students to justify their answers. Bieda and Staples (2020) define justification as “the process of supporting your mathematical claims and choices when solving problems” (p. 103). They argue that justification “is well-known for its role in promoting rigor and developing mathematical understanding. Equally powerful is its role as a discursive practice that provides students with access to mathematical thinking and reasoning (the mathematics behind the answers) and promotes student agency with respect to mathematics” (p. 102). Asking, “Why?” gives agency to the students who are asked to explain their thinking, and it provides access for the students who hear and react to those explanations. When students are asked to justify their answer for this problem, many different explanations may be offered. In a high school class, interpreting an expression could be the key to a proof. The following expression represents the product of 10 consecutive integers: (n)(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)(n + 6)(n + 7)(n + 8)(n + 9) The units digits of those 10 terms will consist of the digits 0 to 9, though not necessarily in that order. For instance, if n = 27, then the units digits will be 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, respectively. This guarantees that the product will be divisible by 100 because the fourth number is a multiple of 10, the sixth number is a multiple of 2, and the ninth number is a multiple of 5. Since 10 × 2 × 5 = 100, the product will be divisible by 100. Another justification is given in the solution below. But more important than multiple proofs are the instructional moves made by a teacher that allow students to explain their thinking and listen to the reasoning of others.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

95

Answer and Solution 10. For a product to be divisible by 100, it must have two 2s and two 5s in its prime factorization. Getting enough 2s is easy; between every pair of 5s, there are two even numbers. But guaranteeing that you have enough 5s is more difficult. Often, a string of six consecutive integers will contain two 5s, such as 15, 16, 17, 18, 19, 20, with the 15 and 20 containing a factor of 5. But to be certain that there are two 5s, it’s necessary to choose at least 10 consecutive integers. As a counterexample, note that the string 6, 7, 8, 9, 10, 11, 12, 13, 14 contains nine numbers but only has one factor of 5. This shows that nine integers are not enough.

PROBLEM 50. Catching Some Zs Common Core State Standards for Mathematics: 8.EE.A.1 • HSS.CP.B.9 • MP.3 • MP.6

In the Classroom The numbers in this problem are too large to be handled by technology. Most calculators will convert to scientific notation at 20! or less. To get students engaged in the context, ask them to find the number of positive integer values of n such that 9!n is an integer. With smaller numbers, 9 investigation via technology becomes possible. Using technology, students will quickly find the values of n for which 9!n is an integer. A class 9 discussion can then occur about why n = 1, 2 are the only solutions. Of course, a simplistic explanation is that 9! = 362,880 is divisible by both 9 and 81. A more sophisticated explanation is that 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 contains the prime factor 3 four times: once in 3, once in 6, and twice in 9. Consequently, 9! is a multiple of 34 = 81 and is therefore divisible by both 9 and 81. Solving and discussing this simpler problem should provide insight for solving the original problem. Note that it’s also possible to have students investigate 10!n , but that expression is almost too similar 10 to the original problem.

Answer and Solution 12 positive integer values. The product 100! = 100 × 99 × 98 × ⋯ × 1. The expression 100n represents a number that has a 1 followed by an even number of zeroes, and it could be rewritten as 102n. Clearly, 100!n will be an integer for n = 1, since 1001 = 100 is the first factor in 100!. Similarly, 100 will be an integer for n = 2, because 1002 = 10,000, and 100! is a multiple of 10,000 since it contains the factors 100, 50, and 2. Likewise, it will be an integer for n = 3, because 100! contains the factors 100, 50, 25, and 8, and 100 × 50 × 25 × 8 = 1,000,000. 96

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

At the heart of this question, then, is how many 2s and 5s appear in the prime factorization of 100!. If we know the answer to that question, we can determine the largest value of n for which 100!  will yield an integer, and every value from 1 to n will be possible. Further, the number of 100 n times that the digit 2 appears in the prime factorization of 100! will be far greater than the number of times that 5 appears; for instance, if you just consider the partial product 100 × 99 × 98 × 97 × 96 × 95, it’s prime factorization is 28 × 33 × 53 × 72 × 11 × 19 × 97, and 2 appears eight times but 5 appears just three times. Consequently, the question really boils down to determining how many 5s appear in the prime factorization of 100!. The number of 5s in the prime factorization of 100! can be easily counted. There are 20 factors that are a multiple of 5, namely 5, 10, 15, . . ., 100; and, there are four factors that are multiples of 25, namely 25, 50, 75, and 100. Therefore, the digit 5 occurs in the prime factorization 20 + 4 = 24 times. Since we need two 5s for 100!n to be an integer, this means an integer value will result for n = 1, 100 2, 3, . . ., 12, so there are 12 values of n for which 100!n will be an integer. 100

PROBLEM 51. Don’t Put Me on a Shelf Common Core State Standards for Mathematics: 2.MD.A.1 • 4.MD.A.2 • 8.EE.C.7 • MP.4

In the Classroom Various versions of this problem have appeared in textbooks and on state assessments for years. The challenge with paper-based versions of this task is that they remove the kinesthetic element and the ability to recognize a linear function in a natural setting is compromised. Moreover, many adaptations of this problem use nice numbers—for example, giving a cup height of exactly 9 centimeters with a lip height of 1.5 centimeters and assuming that all cups are identical—which belies reality. The open-ended presentation given here allows students to engage in a true modeling task. Whether they explore cups provided by the teacher, brought from home, or found at the grocery store or online, students will have to measure and make assumptions about the data collection. For instance, if the cup height varies from 8.8 centimeters to 9.3 centimeters, students will have to make an assumption about average cup height for performing calculations. Making assumptions and deciding how to find that average is a significant component of the modeling process. One interesting presentation of this situation by Andrew Stadel can be found online at http://www. estimation180.com/styrofoamcups.html.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

97

Answer and Solution Approximately 194 cups. To solve this, you’ll need to make some assumptions. Standard Styrofoam cups are about 3.5 inches tall, and there is a rim on each cup that is about 0.5 inch thick. When the cups are stacked, the rim of one cup sits on the rim of the cup below. Consequently, the total height of a stack of n Styrofoam cups is given by the following expression: 3.5 + (n − 1) × 0.5 = 3.0 + 0.5n Since 3.0 + 0.5n = 100 is true for n = 194, it will take 194 cups to reach a height of 100 inches. Of course, Styrofoam cups come in various sizes. Your answer may differ if you consider cups with different dimensions.

PROBLEM 52. Play Ball! Common Core State Standards for Mathematics: 8.EE.C.8 • MP.2

In the Classroom This classic puzzle is a traditional algebra exercise, but it has endured because most people will get it wrong the first time they encounter it. According to Trémolière and De Neys (2014), most people answer incorrectly because of two factors. First, they attempt to simplify the problem, and they unconsciously make the problem easier by removing “more than” and reading that the bat costs $100. Second, the result from this faulty reading delivers a result that is reasonable; it could be the case that a bat costs $100 and a ball costs $10, especially if the person solving it is unfamiliar with the sport of baseball. When presented with the following modified version of the problem, most people answer correctly: A Ferrari and a Ford together cost $190,000. The Ferrari costs $100,000 more than the Ford. How much does the Ford cost?

If a person misreads this problem in the same way, it would give the result that the Ford costs $90,000. That price doesn’t fit with most people’s perception about the price of Ford automobiles, so they’ll read the problem again and make a correction.

98

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Of course, most elementary and middle school students may not have a sense of car prices. Instead, the following modification works well in the classroom: Together, a concert ticket and a CD cost $190. The ticket costs $100 more than the CD. How much did the CD cost?

With an incorrect reading, students might surmise that the CD costs $90. That’s a bit pricey, and it may cause students to reconsider. The correct answer of $45 is still not cheap but more reasonable.

Answer and Solution $105. Algebraically, let t = the cost of the bat and b = the cost of the ball. Then, t + b = 110 and t − b = 100, so 2t = 210, which means that t = 105, and the bat costs $105. Of course, you could also solve this problem without algebra, just using logic. The mean price of the bat and ball is $55. To differ by $100, the bat must be $50 more than the mean, and the ball must be $50 less than the mean. So, the bat costs 55 + 50 = $105.

PROBLEM 53. The Jug Is Half Full Common Core State Standards for Mathematics: 8.EE.C.8 • MP.2

In the Classroom Math teachers are equal parts educator and entertainer. As Butts (1980) noted, “It requires the creativity of an artist to pose a problem so that the potential solver will be motivated to solve the problem” (p. 267). Presenting this problem with a glass vase, colored water, and a scale might provide the motivation that you’re looking for. The vase should be transparent so that students are able to see the liquid within it. Further, it should be either rectangular or cylindrical with a base and sides that have consistent thickness. To prepare for the demonstration, use a marker to draw a line that indicates half the interior height of the vase. When all students have directed their attention to you, fill the vase halfway with colored water; then place the vase on the scale. Give students time to process what you’ve done as well as to note the weight. Then, fill the vase to the top, and again allow students to record the weight. Lift the vase off the scale and dump the water back into the pitcher. Then, hold the vase in one hand as if to pretend that you’re trying to determine its weight.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

99

Answer and Solution 20 ounces. The jug itself has some weight, let’s say j ounces. If the weight of the water is w ounces, then j + w = 100 and j + w2 = 60. Subtracting the second equation from the first gives w2 = 40, which means that w = 80 and j = 20. The weight of the empty jug is 20 ounces. Alternatively, students might deduce that half of the water weighs 100 − 60 = 40 ounces, so all of the water weighs 80 ounces. Therefore, the jug weighs 100 − 80 = 20 ounces. Unfortunately, English is a tricky language. Students might rightfully experience some initial confusion with this problem because the weight of the water is described by the word ounces, yet students (and many adults) often think of ounces as describing the volume of a liquid. Technically, the volume of a liquid should be described by fluid ounces, though colloquial usage is for ounces to indicate both weight and volume.

PROBLEM 54. Dog Day Afternoon Common Core State Standards for Mathematics: 8.EE.C.8.C • HSA.CED.A.1 • MP.4

In the Classroom Displaying only the image of the dogs and the table, ask students what they notice and wonder. Ask students what they know and what they don’t know. The list of things they notice will likely include that the large dog is taller than the small dog; the difference in their heights is either 0 inches or 100 inches, depending on which dog is on the table; and, the height and location of the table don’t change. The list of things they wonder will likely include the height of the table, the height of the large dog, and the height of the small dog. That discussion opens the door for assigning variables to the unknowns in the problem. Then, ask students to generate as many equations about the situation as they can. Possibilities include the following:

100

n

D=T+d

n

D + T = 100 + d

n

D−d=T

n

100 = (D + T) − d

n

d = (D + T) − 100

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Orchestrate a discussion about the equations that were generated. Ask students to state, but not explain, their equations to you. Display a running list of all equations and continue until students have exhausted their lists. Then, call on students to explain the equations that were offered, but not asking any student to explain their own equation. This is a sense-making activity, and there is more value if students are asked to interpret what an equation means instead of describing what they already know. Through the course of this discussion, some students will gain a deeper understanding of the situation, and other students will often realize the errors in their own equations. When all equations have been discussed, you can ask students how these equations could be used to determine any of the unknown heights (small dog, large dog, table). Allow students to discuss this with a partner and, if possible, to solve for any of the unknown heights.

Answer and Solution 50 inches. Let D = the height of the large dog, d = the height of the small dog, and T = the height of the table. The following system of equations represents this situation: D = T + d D + T = 100 + d Since the question does not indicate what we are looking for, we can attempt to solve for any of the variables. One option is to put all the variables on one side of each equation, with constants on the other: D−T−d=0 D + T − d = 100 Adding these equations yields— 2D − 2d = 100 Dividing all terms by 2 gives D − d = 50, so the difference in the height of the dogs is 50 inches. Unfortunately, it’s not possible to find the height of each individual dog. But subtracting the two equations above yields the following: 2T = 100  T = 50 This implies that the height of the table is also 50 inches. Alternatively, logic can be used to find the height of the table. If you combine the left sides of each image, you’ll get a large dog, a small dog, and 100 inches. If you combine the right sides of each image, you’ll get a large dog, a small dog, and two tables. Those amounts will be equal, and if you remove the large dog and small dog from both sides, you’re left with 100 inches on the left and two Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

101

tables on the right. Since those amounts are equivalent, the table is 50 inches tall. (Admittedly, this “logic” is nothing more than algebra in disguise, without the symbolic manipulation.)

PROBLEM 55. The Barnyard Common Core State Standards for Mathematics: 7.EE.B.3 • HSA.CED.A.2 • MP.2

In the Classroom This problem provides an excellent opportunity to employ the three-reads routine, which helps students to carefully read and restate a problem in their own words before attempting to solve it. In the three-reads routine, a problem stem (a math word problem with the question removed; that is, just the context) is read three times, each time with a different focus: 1. Focus on the context; what is happening in this situation? 2. Focus on the mathematics; what quantities are used in this situation? 3. F  ocus on possible problems; what mathematical questions could you ask (and possibly answer) about this situation? For this problem, the first read would involve reading the problem stem without the question “. . . how many of each did she buy?” This allows the student to think about the situation without immediately trying to devise a solution strategy. You could ask, “How would you describe the situation in your own words?” or “Can someone restate this situation using different words?” Prior to the second read, ask students to focus on the numbers and quantities. In this problem, the numbers are 100, 1, 10, 5, and 50, and some quantities are cost and total. Ask students, “What are the quantities in this situation, and how are they related?” Students should be able to explain that some numbers represent costs of individual animals, one represents the number of animals, and the other represents the total cost. During the third read, ask students to consider what mathematical questions could be asked (and possibly answered) about this situation. Emphasize that these should be questions about the quantities and the relationships between those quantities. In this final stage, students will posit many possible problems that could be answered; some possibilities are below:

102

n

How many pigs did she buy?

n

How much did she spend on chickens?

n

Did she spend more on cows or pigs?

n

Did she buy more pigs or chickens?

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

It’s quite likely that students will generate questions that are more difficult to answer than the one included in the problem! Give students time to answer any of the questions that were posed. If the question included in the problem—how many of each did she buy—was not suggested by students, ask them to answer that question in addition to the ones they selected. Using the three-reads routine allows a problem to be scaffolded so that students can make sense of the problem at their own pace and in their own way. Consequently, it helps to reduce anxiety. Often, students will solve the question that appears in the problem before it is even asked.

Answer and Solution 1 cow, 9 pigs, 90 chickens. If there are c cows, p pigs, and k chickens, the following equations result: c + p + k = 100 10c + 5p + 0.5k = 100 Doubling the second equation and switching the order gives— 20c + 10p + k = 200 c + p + k = 100 Subtracting the second equation from the first then yields 19c + 9p = 100. This is an equation with two variables, which has an infinite number of solutions. But since the answers must be limited to integers—it’s not possible to have 23 of a cow, for example—the only possible solution is c = 1, p = 9. Then, c + p + k = 1 + 9 + k = 100, and k = 90.

PROBLEM 56. Sum of Cubes Common Core State Standards for Mathematics: HSA.SSE.A.1.B • HSN.RN.A.2 • MP.2

In the Classroom All students learn to solve algebraic equations as part of their mathematical education, but what do you do when traditional methods won’t work? This problem involves one equation with three unknowns. One informal definition of problem solving is “knowing what to do when you don’t know what to do.” Without very advanced techniques, a systematic approach is the only way to solve this problem. One by one, replace z with 1, 2, 3, . . ., until something good happens. Fortunately, something good happens when z = 7, as the equation reduces to x3 + y3 = 100 − 73, or x3 + y3 = −243. An integer solution to that equation is x = −3 and y = −6. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

103

Answer and Solution {−3, −6, 7}. Originally posed in 1954 at the University of Cambridge, the Sum of Three Cubes problem asked for solutions to the Diophantine equation x3 + y3 + z3 = k where k is an integer and 1 < k < 100. Over the course of several years, solutions were found for all values of k except 33 and 42; it wasn’t until 2019, with the help of shared processing power borrowed from computers around the globe, that solutions for those two numbers were found. The given problem, with a sum of 100, is much easier than those with sums of 33 or 42. By looking at the cubes of the single-digit integers, you’ll notice that 73 is 100 more than 33 + 63. 13 = 1 23 = 8 33 = 27 43 = 64 53 = 125

63 = 216 73 = 343 83 = 512 93 = 729

Once you have that, it’s easy enough to realize that (−3)3 + (−6)3 + 73 = 100.

PROBLEM 57. Absolute Power Common Core State Standards for Mathematics: 6.NS.C.7.C • HSA.REI.D.11 • HSF.IF.C.7 • MP.5

In the Classroom Graphing the function will not reveal all of the integer solutions, but it will provide insight regarding the relationship between x and y as well as the maximum possible values of each. Students should be encouraged to use a graphing calculator to help understand the situation. Students can also create a table of values by hand, use a spreadsheet to efficiently determine a complete solution, or use logic to identify the four vertices of the graph. In middle school, students learn that the absolute value of a number is “its distance from 0 on the number line” (CCSSI 2020, p. 43), and the analog to this problem is that the sum of |x| + |y| represents the combined distance from the x- and y-axes. Interestingly, the following extension contains only a minor modification from the original problem, but the solution is very different: If x and y are both integers, how many different solutions exist for |x + y| = 100? The graph of this function is two parallel lines which extend infinitely in either direction. Consequently, there are infinite solutions to the equation. 104

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

500

0

–500

500

–500

Answer and Solution 400. The following graph of the function yields a lot of insight about the possible integer solutions: 100

50

–100

–50

0

50

100

–50

–100

If −100 < x < 100, then y = ±(100 − |x|). But when x = ±100, the only possibility is y = 0. Consequently, there are 2 × 199 + 2 = 400 solutions for which x and y are both integers.

PROBLEM 58. Absolutely Dreadful Common Core State Standards for Mathematics: 6.NS.C.7.C • 7.NS.A.1.C • HSA.REI.D.11 • MP.7

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

105

In the Classroom It is absolutely possible to graph this function and determine the minimum value by visual inspection. Unfortunately, graphing the function also reduces the cognitive load and removes the opportunity for some algebraic interpretation, especially if technology is used to generate the graph. Hawthorne and Druken (2019) identify “decomposing (or chunking) algebraic expressions into a variety of substructures based on the context and goals” (p. 298) as one of the components of structural reasoning. As presented, the expression to the right of the equals sign is daunting, but it is worth asking students whether there is a smaller portion that they can deal with. Students who understand absolute value should be able to analyze just |x − 100| and realize that the minimum value of that expression is 0 when x = 100. When the next part of the expression is included to yield ||x − 100| + 100|, students should recognize that the addition of + 100 will increase the maximum from 0 to 100. And so on. By helping students to attend to portions of the problem instead of the entire function at once, it reinforces the strategy of “solve a simpler problem” and, in the process, helps to reduce student anxiety.

Answer and Solution 100. One possibility is to construct a table of values and plot some points. Let’s take a look and see what happens. x

y

–50

250

0

200

50

150

100

100

150

150

200

200

250

250

Plotting these points suggests that the minimum value of y = 100 occurs when x = 100. A close examination of the function will help to explain why this happens. The first part of the expression on the right-hand side, |||x − 100| + 100| − 100|, is completely contained within absolute value symbols. Consequently, the value for this portion of the expression can never be negative. But, can it ever be zero? Yes, when x = 100. Therefore, the least possible value of y = 100 is obtained when x = 100. 106

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 59. Don’t Be Mean Common Core State Standards for Mathematics: 6.SP.A.3 • HSS.ID.A.2 • MP.1 • MP.5

In the Classroom Get messy! There is no way to understand mathematics without getting into the nitty-gritty, so students should be encouraged to generate a set for which the problem statement is true. Of course, a set with 100 elements is a bit unwieldy, so students should attempt to generate a smaller set—for instance, a set of eight numbers such that the average of the first seven numbers equals the average of the entire set. This will yield insight into the nature of the mean. Students can use a spreadsheet to generate such a set.

Answer and Solution The 100th number is equal to the mean of the entire set. If m is the mean of the first 99 numbers, then the sum of the first 99 numbers is 99m. Assume that n is the value of the 100th number. Because the mean of all 100 numbers is also m, then 99m + n = m 100 99m + n = 100m n = m

Therefore, the 100th number in the list must be equal to the mean.

PROBLEM 60. Counterintuitive Cards Common Core State Standards for Mathematics: 7.SP.A.2 • 7.SP.C.6 • HSS.MD.B.5.A • MP.3

In the Classroom Students will initially believe that choosing the opposite color is the correct strategy, but a simulation will reveal that this is not the case. To reduce the preparation, instead of creating a deck with 100 cards, the game can be simulated using 10 cards: 3 red-red, 3 blue-blue, and 4 red-blue. Even after 10 or 15 trials, students often continue to believe that the strategy of choosing the opposite color is correct. They’ll insist that they’ve just been unlucky. This allows for a discussion of how many trials are necessary to have confidence that the results are accurate and not an anomaly. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

107

For a 95 percent confidence level with a 5 percent margin of error, more than 300 trials would be needed; for an 80 percent confidence level with a 10 percent margin of error, fewer than 50 trials would be needed. Often, a statement as simple as “Let me know when you’re convinced” will be sufficient for students to acquiesce. The purpose of ceasing the experiment, however, is not to ensure that students agree with you; rather it’s so that students can start to explore why the different-color strategy is incorrect or to explain why the same-color strategy is optimal.

Answer and Solution Whatever color the dealer shows you, predict that the other side will be the same color. This strategy seems counterintuitive, but probability can show that it really is the best tactic. Without loss of generality, let’s say that the dealer shows you a card with a red side. There are 30 cards that have red on both sides but 40 that have red on one side and blue on the other, and since 40 > 30, it’s typical to think that it’s more likely for the other side to be blue. It would seem that 40 4 = . Unfortunately, that thinking is incorrect. P(blue) = 70 7 In truth, when the dealer shows you the red side of a card, it could be the red side of a red-blue card, or it could be either side of a red-red card. Because there are 30 red-red cards, the dealer could be showing you any one of the 60 red sides that have red on the other side, too. Consequently, 4 40 2 P(blue) = 100 = , not . The odds are strongly in favor of the same color occurring, so the optimal 7 5 strategy is to choose the same color that the dealer shows you. While it’s possible to predict the long-term outcome of this game, it’s not possible to accurately predict the outcome on any given turn. Consequently, this strategy only works if you’re willing to play long enough for the law of averages to take effect.

PROBLEM 61. Tickets, Please Common Core State Standards for Mathematics: 7.SP.C.6 • HSS.MD.B.7 • MP.3

In the Classroom “Act it out” is a valuable strategy that allows students to “see” the mathematics without attempting to articulate a formal solution. For this problem, students can act it out in groups of various size. Groups of five work well, with the numbers 1 to 5 written one each on a piece of paper; these represent the “tickets,” and they should be shuffled and randomly distributed to the students in the group. Similarly, five desks should be numbered 1 to 5 to represent the seats on the train. Then, 108

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

without looking at the number, one student should set her ticket face down on a desk. Students can then act out the scenario and see what happens. On the board or a piece of chart paper, create a table as follows: Number of Seats/ Tickets

Last Person Got Their Assigned Seat

Last Person Did Not Get Their Assigned Seat

2 3 4 5 6

The first column can be filled with the number used in various trials, and any number can be used. Tally marks can then be used to record the number of successful and unsuccessful attempts. Every group can record their attempts in the same table to gather a larger collection of data. Giving each group a marker of a different color will make it easier to identify outliers or, in some cases, to remove or modify incorrectly recorded data. As the data begins to grow, it will start to become clear that the odds of the last person getting their assigned seat is 50:50. What becomes difficult is explaining why that’s the case, and the best way to understand it is to act out the problem a few more times and pay attention to what happens. Students will naturally gravitate to the explanation provided at the end of the solution below.

Answer and Solution 1 2 . The number 100 in this problem is a red herring. The probability of getting your assigned seat is

the same whether there are 2 seats, 10 seats, or 100 seats.

As the simplest example, consider a train with just two seats. The probability that the first person randomly chooses her assigned seat is 12, which means that the probability that you get your assigned seat is also 12. With three seats, the situation is a little more complex but still manageable. To make this easy, let’s assume the first person was assigned Seat 1, the second person was assigned Seat 2, and the third person (you) was assigned Seat 3. There are three scenarios to consider: 1. O  ne-third of the time, the first person will randomly, and correctly, choose Seat 1. Then the second person will get Seat 2, and Seat 3 will be unoccupied when you board. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

109

2. O  ne-third of the time, the first person will randomly choose Seat 2. That means that the second person will choose a seat at random, half the time choosing Seat 1, in which case Seat 3 will be unoccupied when you board, and half the time choosing your seat. 3. A  nd one-third of the time, the first person will choose Seat 3. That’s your seat! So, there is no chance that your assigned seat will be unoccupied if that happens. Putting this all together, the probability that your seat will be unoccupied when you board is 1 × 1 + 13 × 12 + 13 × 0 = 13 + 61 = 12 . 3 As it turns out, the probability is 12 no matter how many seats are on the train. You can check it for four, five, or a thousand seats, but here’s why it works that way. Half the time, everything will sort itself out, meaning that someone randomly chooses the first person’s seat; therefore, everyone thereafter can sit in their assigned seat, leaving your seat unoccupied. And half the time, someone will randomly choose your seat, so you’ll be forced to sit somewhere else. An alternative explanation is that your fate—that is, whether you get your seat or not—will be determined when anyone who boards selects either your seat or the first person’s seat. If a passenger takes any other seat, that doesn’t resolve the situation. As each person boards, either the person selects an inconsequential seat (possibly hers, or possibly another seat that is neither the first person’s seat nor your seat), or there is a 50 percent chance that the situation gets settled either way. The situation can be resolved by any passenger whose seat has been taken, and if it is resolved, the probability is 1  that that passenger will select the first person’s seat so that yours will remain available. 2

PROBLEM 62. You Say It’s Your Birthday Common Core State Standards for Mathematics: 7.SP.C.8 • HSS.CP.A.1 • MP.4

In the Classroom Originally posed by Richard von Mises in 1939 (von Mises and Birkhoff, 1964), the Birthday Problem has been explored by many people since. Of note, the term birthday in this scenario actually refers to a birth date, which is its common colloquial usage. This problem can be presented as a data collection and exploration activity. In most classrooms, it may not be possible to collect birthday data for 100 people. With some advanced planning, data could be collected either from students in multiple classrooms, students could supply the birthdays of members of their families, or a data set with birthdays of famous people could be generated with some internet research.

110

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

With data available, ask the class, “How many people would need to be in a room so it’s likely that two of them share the same birthday?” A brief discussion about likely can occur, since this term can often have different interpretations. Maubossin and Maubossin (2018) found that there is a wide discrepancy between how different people interpret the same probabilistic word; most respondents in their survey generally associated the word likely with a probability of anywhere from 40 percent to 90 percent. For the purpose of this investigation, the class should agree on what likely means by assigning it a percentage. Using a data set with a large number of birthdays, the experimental probability can be found for groups of various size. Working in pairs, students can randomly select (using a random number generator) a group of 10 people from the list and see whether any of them share a birthday. The probability is about 0.11, so a few pairs of students may find a set with two people who share a birthday. That alone will be a revelation to the class. The same activity can be repeated with pairs randomly choosing groups of 15, 20, and 25 people, or more. A record of how many times the random groups did and did not contain a shared birthday should be noted, and the experimental probability can be computed. Students will start to notice that shared birthdays become likely when the group size grows to about 25 people. In fact, with just 23 people in a room, the likelihood of a shared birthday is greater than 50 percent. The experimental results typically provide impetus for students wanting to determine the theoretical probability of a shared birthday with 100 people.

Answer and Solution 99.999977 percent. Calculating the probability that two people have the same birthday is rather 1 chance that Person B has the same birthday. Not easy. Whatever Person A’s birthday is, there is a 366 very likely. (Assume that all dates are equally likely, even though that’s not true.) Calculating the probability that three people have the same birthday is a little more difficult because you have to consider all three combinations in which two of them share a birthday as well as the possibility that all three share the same birthday. But calculating the probability that they do not have the same birthday is a little easier. Whatever Person A’s birthday is, there 365 chance that Person B doesn’t have the same birthday. Moreover, there is a 364 chance that is a 366 366 Person C doesn’t have the same birthday as either Person A or Person B. Therefore, the probability that (at least) two of them share a birthday is—  365   364  × = 1 − 0.992 = 0.008 1−  366   366  Again, not very likely.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

111

But if there were 100 people, then the probability that at least two of them share a birthday is much more likely. In fact, it works out to—  365   364   363   266  × × ×⋯× = 1 − 0.00000023 = 0.99999977 1−    366 366 366        366  —a virtual guarantee.

PROBLEM 63. The Envelope, Please Common Core State Standards for Mathematics: 7.SP.C.6 • HSS.CP.A.3 • HSS.IC.B.5 • MP.4

In the Classroom The situation described in this beautiful problem was presented by Paulos (1998) in Innumeracy: Mathematical Illiteracy and Its Consequences as an example of how the population often places an overreliance on averages. For that very reason, simulating this problem in the classroom can lead to powerful insight and rich discussions. The important aspect of the problem is that the amount of $100 is completely random, and it serves as a red herring during analysis. To conduct a simulation in class, grab a large collection of fake money from a board game and two envelopes. With your back to students, fill the envelopes with two amounts of money such that one amount is double the other. To ensure that you aren’t inadvertently displaying tendencies, a random number generator should be used to pick the amounts. Moreover, the amount of money to be placed in each envelope should be even; an odd amount might lead students to suspect than the other envelope couldn’t contain half as much, especially if coins are not used. A Google sheet that helps with this simulation can be found at https://tinyurl.com/TwoEnvelopes. The class can run multiple simulations and keep track of whether switching was beneficial. With enough trials, it will become clear that switching doesn’t make sense.

Answer and Solution Keep the $100. The envelope you chose has $100 in it, so it would seem that the other envelope must have either $50 or $200. Hence, switching seems like the right thing to do as you could increase your amount by $100 but only decrease your amount by $50. The problem is that there isn’t a 50 percent chance that the other envelope has $50 or a 50 percent chance that it has $200. Rather, there is either a 100 percent chance the other envelope has $50 and a 0 percent chance that it has $200, or there is a 0 percent chance the other envelope has $50 and a 100 percent chance that it has 112

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

$200. That is, the amount in the two envelopes was decided before you made your selection, not after. As Paulos (1998) explains: The question is: Why didn’t you choose [the other envelope] in the first place? It’s clear that no matter what amount of money was in the envelope originally chosen, given permission to change your mind, you would always do so and take the other envelope. Without any knowledge of the probability of there being various amounts of money in the envelopes, there is no way out of this impasse. Variations of it account for some of the “grass is always greener” mentality that frequently accompanies the release of statistics on income. (p. 127)

Unfortunately, that explanation from Paulos is insufficient to understand why switching is inconsequential. Instead, consider the following, and slightly more compelling, explanation from Thornton (personal communication, February 5, 2020): The answer depends on what is meant by “should.” If the objective is to maximize the expected take, then switching doesn’t make a bit of difference. Suppose that one envelope contains x dollars and the other contains 2x dollars. With a random choice, the expected take is 1.5x. Further, even if the decision has been made ahead of time to reverse the original choice, that doesn’t change the fact that it’s a random choice between two envelopes. The expected take is still 1.5x. Knowing the contents of the first envelope is a red herring; it doesn’t reveal the value of x, and thus doesn’t change the expected outcome. On the other hand, suppose you have an immediate and critical need for something that costs $200. Then you “should” change, because you might get the money you need. 

That explanation from Thornton seems like it would be better received by the average high school student.

PROBLEM 64. Wrecked Angle Common Core State Standards for Mathematics: 4.MD.A.3 • MP.8

In the Classroom An important part of posing problems in the classroom is ensuring that all students have an entry point. For some students, the prospect of finding every rectangle with a perimeter of 100 units Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

113

could be a formidable task. So instead, ask students to find just one rectangle whose perimeter is 100 units. Allow students to share their answers with a classmate. In some cases, students will immediately realize that they found a rectangle with an area, not a perimeter, of 100 square units; in other cases, students will realize that they found a rectangle for which the sum of the length and width is 100 units, which means the rectangle would have a perimeter twice as large as it should be. A discussion with a classmate will often allow them to recognize their mistake and make a correction. After sharing with just one classmate, students can share with the entire class. This provides an opportunity to generate a collective list. Once several different rectangles have been suggested, the class can be turned loose on the full problem. Pólya (1945) said, “If you can’t solve a problem, then there is an easier problem you can solve: Find it.” Presenting the problem in this way allows all students to find an easier problem that they can solve.

Answer and Solution 25 rectangles. If the perimeter is 100 units, then the sum of the length and width must be 50 units. The dimensions could be 1 × 49, 2 × 48, 3 × 47, . . ., 25 × 25.

PROBLEM 65. Area of Influence Common Core State Standards for Mathematics: 4.MD.A.3 • MP.8

In the Classroom As with problem 64, Wrecked Angles, this task lends itself to finding just one rectangle with the given area first. Attempting to find the entire list proves relatively straightforward for most students, though many will overlook the obvious 1 × 100 rectangle. The benefit of using these two problems, however, lies in coupling their collective power. The corollary questions “What rectangle with integer dimensions and an area of 100 square units has the greatest (or least) perimeter?” and “What rectangle with integer dimensions and a perimeter of 100 units has the greatest (or least) area?” allow for playful exploration and encourage students to investigate some sophisticated mathematics. Both of these problems can serve as the basis for an entire lesson. The latter problem, in particular, requires students to come face-to-face with a quadratic relationship because the dimensions will be x and 50 − x, and the area will be A = x(50 − x) = 50x − x2. Young students won’t use this notation, but exploration will allow them to realize that a square is the rectangle with greatest area for a given perimeter.

114

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Answer and Solution Five rectangles. To have an area of 100 square units, the product of the length and width must be 100. The dimensions could be 1 × 100, 2 × 50, 4 × 25, 5 × 20, or 10 × 10.

PROBLEM 66. Around the Squares Common Core State Standards for Mathematics: 3.MD.D.8 • MP.3

In the Classroom Before presenting this problem to students, display the image without the problem and ask, “What do you notice?” Student responses will range from mundane (there are squares; there are four “tails”; all squares are congruent) to practical (there are 100 squares; adjacent squares are connected along a side) to mathematically sophisticated (the “center” square is completely surrounded; the squares at the end of each tail contribute three units to the perimeter; all other squares contribute two units to the perimeter). Student responses might also include questions, such as “How many squares are there?” and “Are all of the shapes actually squares?” Even though you haven’t shared the problem with them, it would be okay to confirm these characteristics about it as the intent is not to have students count the squares one by one or to prove that they are squares. You can also ask students, “What do you wonder?” Students might wonder about the area of the figure, the perimeter of the figure, or other aspects. Allowing students to make observations about a problem before solving it gives them agency and reduces anxiety (Tobias 1993). Moreover, allowing students to choose which question(s) they’d like to answer helps to build a problem-solving culture in your classroom because it empowers students and allows for differentiation as students naturally attempt problems that they think they can handle. If all students elect to find the perimeter, a whole-class discussion can follow. Ask, “Give an estimate for the perimeter that you know is too high,” and “Give an estimate for the perimeter that you know is too low.” Push students to give estimates that include mathematical thinking, not just guesses. For instance, a student might suggest, “There are 100 squares, and each square has four sides, but not all of the sides are exposed. The perimeter will be less than 4 × 100 = 400 units.” But if students opted to answer different questions, it would be appropriate to say, “If you haven’t already determined the perimeter of the figure, please solve this problem, too,” and display the full problem.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

115

Answer and Solution 202 units. Most of the squares are connected to squares along two of their sides, which means that the other two sides contribute to the perimeter. The five squares that are not connected to two other squares are shaded in the figure below: The four red squares are only connected to one other square, so they contribute three units to the perimeter, while the black square is connected to squares on each of its four sides, so it contributes nothing to the figure’s overall perimeter.

The total perimeter is 95(2) + 4(3) + 1(0) = 202 units.

PROBLEM 67. Chains of Fools Common Core State Standards for Mathematics: 4.G.A.2 • 5.OA.B.3 • 8.F.A.3 • 8.F.B.4 • MP.4

In the Classroom The Common Core State Standards for Mathematics requires students to analyze patterns in upper elementary school and explore linear functions in middle school. This problem spans that conceptual development. By exploring the chain formed by each shape, students will notice a number pattern. For instance, the perimeters for 1, 2, 3, . . . pentagons will be 5, 8, 11, . . . units, a linear pattern that increases by 3 units from term to term. This pattern can be generalized by the linear function P(n) = 3n + 2, where the rate of change represents the number of units by which the perimeter increases at each step and the y-intercept accounts for the extra side that occurs on the first and last pentagons. In middle school, students are expected to solve two-step linear equations, and setting this function equal to 100 provides a context in which to solve the equation 3x + 2 = 100. That equation is true for 2 x = 32  , and students can interpret the meaning of the solution in context. Because the solution is 3 not an integer, a chain of pentagons cannot have a perimeter of exactly 100 units.

116

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

This problem can be extended and approached as a class problem-solving task. Collectively, students could attempt to determine the regular polygons for which it would be possible to form a chain with a perimeter of 100 units. Every student in class could choose a shape to investigate, and the results could be shared first in small groups, and then in a whole-class discussion. Eventually, the class should come to consensus about which shapes are possible, and they should be able to provide a justification for their answer.

Answer and Solution Only squares can be extended to have a perimeter of 100 units; pentagons, hexagons, heptagons, and octagons cannot. The total perimeter of each chain depends on the shape. For n shapes, the perimeter of each chain is: n

Squares: 2n + 2

n

Pentagons: 3n + 2

n

Hexagons: 4n + 2

n

Heptagons: 5n + 2

n

Octagons: 6n + 2

To see which chain can have a perimeter of 100 units, set each of those expressions equal to 100, and find all of those that yield an integer value for n. n 2n + 2 = 100 → 2n = 98, n = 49 n 3n + 2 = 100 → 3n = 98, n = 32.7 n 4n + 2 = 100 → 4n = 98, n = 24.5 n 5n + 2 = 100 → 5n = 98, n = 19.6 n 6n + 2 = 100 → 6n = 98, n = 16.3

Because the expression for squares is the only one that yields an integer value for n, the square chain is the only one that can have a perimeter of 100 units.

PROBLEM 68. Rock the Octagon Common Core State Standards for Mathematics: 3.MD.D.8 • 4.MD.A.3 • 5.G.B.3 • MP.4

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

117

In the Classroom At its most basic level, this problem involves area and perimeter, which are topics addressed in the Common Core State Standards for Mathematics for grades 3−5. But finding its solution requires problem-solving skills beyond those of a typical elementary student. The first question that often arises for students is “Is it even possible to create an octagon with the given side lengths, so that all of the sides meet at right angles?” That’s a fair question, and it’s one worth exploring—without the encumbrance of worrying about finding the octagon with greatest area. Using graph paper, students can attempt to create an octagon with side lengths of 9, 10, 11, 12, 13, 14, 15, and 16 units. When students have constructed an octagon with the given lengths and perpendicular sides, they can then decompose the figure to determine its area. One efficient method for finding the area is placing a circumscribed rectangle around the octagon, and then subtracting the area of any small rectangles that are within the rectangle but outside the octagon. After the area is found, the natural question is “Is it possible to make a larger octagon?” If a circumscribed rectangle has been used to find the area, that may lend insight to a process for finding the largest octagon. What is the area of the largest rectangle that could be circumscribed about the octagon? What is the area of the smallest rectangles that could be removed? To find the solution for this problem, students have to complete three steps: (1) create an octagon with the given side lengths, which is a nontraditional problem-solving task; (2) determine the area of the octagon they’ve created, which is a typical elementary grades exercise; and (3) determine whether a larger octagon is possible, which requires students to provide justification. Taken together, these three steps provide an incredible problem-solving opportunity for students. Reiter, Holshouser, and Vennebush (2012) authored an article about lattice octagons for the Mathematics Teacher that provides extension problems and additional suggestions for classroom use.

Answer and Solution 407 square units. This problem lends itself to the application of perhaps the most fun problemsolving strategy: “wishful thinking,” in which you modify the problem to be one that is less overwhelming in the hopes that it will provide insight for the more difficult problem that you’ve actually been asked to solve. The given problem asks for an octagon to be created from the eight given side lengths, and that seems like it might be hard to do. But what if we could create a different shape, say, a rectangle? Moreover, since the problem asks for the maximum possible area, what if we could create a square—since it’s the quadrilateral with the greatest area for a given perimeter?

118

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

When employing wishful thinking, it’s not always clear what will happen. It’s a strategy to use when progress on the initial problem has been limited. It might not lead to a solution; then again, it might, and in the meantime, it gives you something more productive to do that just sitting and waiting for inspiration to arrive. Can we make a square with a perimeter of 100 units? Yes, and each side of the square must be 25 units. Consequently, two of the given side lengths will be needed for each side of the square, and one possibility is shown below.

Does that square provide any information about how to produce an octagon? In fact, it does. Imagine pushing opposite corners so they fold into the middle. The result will be an octagon with adjacent sides perpendicular and with the required side lengths. The shaded region shown below is one possible octagon:

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

119

The question, of course, is whether this is the octagon with maximum size. In fact, it’s not. But finding its area will provide further insight. The area of the square that surrounds it is 252 = 625 square units, and the two unshaded rectangles are 9 × 13 = 107 square units and 11 × 15 = 165 square units, respectively. Consequently, the area of the shaded octagon is 625 − 107 − 165 = 353 square units. Is it possible to create a larger octagon? Or said a different way, is it possible to remove less area from the two corners that were collapsed? The answer is yes. The four smallest given side lengths were 9, 10, 11, and 12, and if they can be used to form the removed rectangles, the area of the resulting octagon will be larger. In fact, the best arrangement is as follows:

As a result, the maximum area is 252 − (9 × 12) − (10 × 11) = 625 − 108 − 110 = 407 square units.

PROBLEM 69. Flattening a Triangle Common Core State Standards for Mathematics: 6.G.A.1 • HSN.Q.A.3 • HSG.MG.A.3 • MP.5

In the Classroom This problem lends itself to the use of technology in two different ways. One option is to generate a function that expresses the area of the triangle in terms of the base, b, and then graph that function. An alternative option is to create a diagram with dynamic geometry software that can be used to explore the situation. As shown in the solution below, the area function is A(b) = 1 × b × 1002 − 0.25b2 . The graph of 2 this function can be seen at https://www.desmos.com/calculator/ssbpgir1y1. The maximum value

120

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

of this function occurs when b ≈ 141.421. Students may recognize the relationship of this number to the decimal value of 2. That recognition may spark the realization that the maximum area occurs when the angle between the two legs is a right angle; if both legs are 100 inches, the hypotenuse will be 100 2 ≈ 141.421. An interactive geometry simulation for this situation is available at https://www.desmos.com/ geometry/qagftur4m5. The orange dot at the top of the triangle can be pushed and pulled vertically to change the height of the triangle; the area is indicated inside the polygon. Unfortunately, it’s not possible to make a side length of 100 inches, so the legs of this triangle measure 10 units. Through exploration, students will see that the maximum area is approximately 50 square units; converting to inches and replacing 10 units with 100 inches reveals the correct answer: 5,000 square inches. If the graph of the function is used to determine the maximum area, it should not be shared with students until they have correctly determined the area function. Similarly, if the geometry construction is shared with students, manipulating the figure to find the maximum area is insufficient. Push students to provide a justification for any discoveries they make. (The original problem, Sliding Triangle (Niederman 2009), appears on the NCTM Illuminations website.) https://www.nctm.org/Classroom-Resources/Problems-(Brain-Teasers)/Sliding-Triangle/

Answer and Solution 5,000 square units. If the two congruent sides are vertical, the triangle will be a degenerate one with a base length of 0 units. If the top vertex is pushed until the triangle is flattened, the triangle will be degenerate with a base length of 200 units. What happens between those two extremes is worth investigating. When the top vertex is pushed so that the base length is b units, the height of the triangle will be  1 h = 1002 − 0.25b2, and the area of the triangle will be A = × b × 1002 − 0.25b2 . 2

The following table shows the area of the triangle as the base length b increases from 0 to 200.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

121

Base Length

Height

Area

0

100.0

0.0

10

99.9

499.4

20

99.5

995.0

30

98.9

1483.0

40

98.0

1959.6

50

96.8

2420.6

60

95.4

2861.8

70

93.7

3278.6

80

91.7

3666.1

90

89.3

4018.6

100

86.6

4330.1

110

83.5

4593.4

120

80.0

4800.0

130

76.0

4939.6

140

71.4

4999.0

150

66.1

4960.8

160

60.0

4800.0

170

52.7

4477.7

180

43.6

3923.0

190

31.2

2966.4

200

0.0

0.0

Inspecting the table suggests that the maximum area of approximately 4,999 square units occurs when b = 140 units. 1

To find the exact value, though, it’s possible to graph the area function A = × b × 1002 − 0.25b2 . 2 The graph below, created at www.desmos.com, shows that the maximum area of 5,000 square units occurs when b ≈ 141.4 units.

122

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

(141.421, 5000) 5000

0

200

This result makes sense. A square is the quadrilateral with maximum possible area for a given perimeter. Using similar logic, the maximum possible area of this flattened triangle will occur when the angle between the two legs is a right angle. At that point, the area of the triangle will be equal to one-half the product of the congruent legs, so the maximum area is 12 × 100 × 100 = 5,000 square units.

PROBLEM 70. Pieces of Ten Common Core State Standards for Mathematics: HSG.SRT.B.5 • MP.6

In the Classroom Since this problem already contains a figure, the problem-solving strategy “draw a picture” is not completely applicable. But adding elements to the figure might prove helpful. It’s worth giving students several minutes to think about what line segments they might add to the figure. Then, allow them to discuss their thoughts with a partner before attempting to solve the problem. Conduct a discussion about which segments to add and why they should be added. Students may suggest quite a few, and the class can discuss the relative merits of each. Segments that connect the center to a vertex are often helpful; segments that divide shapes into congruent pieces can be useful, too. Moreover, cutting and moving pieces can be effective in that it can be used to show that pieces have the same area or that some pieces can be combined to produce more advantageous shapes. The solution below shows one way that adding segments and rearranging pieces can lead to a correct result.

Answer and Solution 20 square units. This surprising but rather satisfying result can be obtained without trigonometry or assuming lengths just by using geometrical logic.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

123

The decagon is divided into eight triangles; due to symmetry, there are four pairs of congruent triangles. Let’s start with the two largest triangles, which happen to be right triangles. The midpoint of their shared side is also the center of the decagon. From the center, draw a segment to a vertex at the right angle of one of the triangles, as shown:

This drawn segment separates the large triangle into two equal halves. Moreover, if similar lines were drawn from the center to every vertex, they would divide the decagon into ten congruent triangles. Therefore, the newly formed triangle shown above, whose base is a side of the decagon, has an area equal to one-tenth of the entire decagon. Since the two triangles into which the large right triangle was divided have equal area, the total area of the large right triangle is equal to one-fifth the area of the decagon, or 20 square units. Now consider one of the two smallest triangles into which the decagon was divided. Then, draw a segment that divides this triangle into two smaller right triangles. If those triangles were then moved and attached to the bottom of one of the second-largest triangles as shown below, they would collectively form a right triangle that is congruent to the two large right triangles in the original decagon. Consequently, this combination of triangles has an area of 20 square units, also.

With all of that hard work out of the way, the only remaining part is to state the punch line. The combined area of all the unshaded regions is 80 square units, so the combined area of the two shaded regions is 20 square units. Further, due to symmetry, each of the shaded regions has an area of 10 square units. It should be noted that the succinct description of the solution above is the result of several hours of problem solving. What’s not evident are the false starts and missteps that were part of the process. The brevity of the solution should not suggest that this problem will be solved quickly. 124

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

PROBLEM 71. Prism to the Max Common Core State Standards for Mathematics: 5.MD.C.3.B • 5.MD.C.5.B • 6.G.A.2 • 7.G.B.6 • MP.3

In the Classroom A set of 100 omnifix cubes or wooden cubes can be used to support students who have not yet developed the spatial visualization skills to solve this problem. Likewise, the use of those manipulatives will allow students to see the structure of mathematics because prisms with the given volume will have dimensions that are factors of 100. This may not be an observation that students make on their own, so a question such as “What do you notice about the dimensions of the prisms that you’ve found?” might instigate a student to look for a pattern or similarities in the numbers.

Answer and Solution Eight rectangular prisms. The prime factorization of 100 is 22 × 52. Consequently, a prism with a volume of 100 cubic units will have dimensions that are multiples of 2, of 5, or of both. There are eight such prisms: 1 × 1 × 100, 1 × 2 × 50, 1 × 4 × 25, 1 × 5 × 20, 1 × 10 × 10, 2 × 2 × 25, 2 × 5 × 10, and 4 × 5 × 5.

PROBLEM 72. Run for Cover Common Core State Standards for Mathematics: 7.G.B.6 • MP.4

In the Classroom For most students, the difficulty of this problem lies not with the calculations but with the ability to visualize the different prisms that are possible. A set of 100 omnifix cubes or wooden blocks can be used to model the various prisms that can be created. As a whole-class discussion, identify all of the prisms that are possible and list their dimensions on the board or chart paper. Students can then work in pairs or groups of three to assemble the prisms and determine their surface area. It’s quite likely that after constructing just a few prisms, students will see that the surface area is smaller when the dimensions are close to one another, and the surface area is greater when the dimensions are less similar. Consequently, students likely won’t have to build all eight prisms to determine the answer.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

125

Answer and Solution 402 square centimeters. As shown in the table below, of all rectangular prisms whose volume is 100 cubic centimeters, the prism measuring 1 × 1 × 100 has the greatest surface area. Dimensions

Surface Area

1 × 1 × 100

2 × (1 × 1 + 1 × 100 + 1 × 100) = 402

1 × 2 × 50

2 × (1 × 2 + 1 × 50 + 2 × 50) = 304

1 × 4 × 25

2 × (1 × 4 + 1 × 25 + 4 × 25) = 258

1 × 5 × 20

2 × (1 × 5 + 1 × 20 + 5 × 20) = 250

1 × 10 × 10

2 × (1 × 10 + 1 × 10 + 10 × 10) = 240

2 × 2 × 25

2 × (2 × 2 + 2 × 25 + 2 × 25) = 208

2 × 5 × 10

2 × (2 × 5 + 2 × 10 + 5 × 10) = 150

4×5×5

2 × (4 × 5 + 4 × 5 + 5 × 5) = 130

Of note, the prism that most closely resembles a cube (4 × 5 × 5) has the least surface area.

PROBLEM 73. Strike a Chord Common Core State Standards for Mathematics: 7.G.B.4 • HSG.C.A.2 • MP.3 • MP.4

In the Classroom The solution given below relies on algebra, but wishful thinking can be an effective strategy to use with this problem. That said, students should have the opportunity to solve the problem using any method they like before orchestrating a discussion of particular strategies. The only dimension given in the problem is 100 centimeters for the length of the chord; the radii of the two concentric circles are unknown. To convince students that the radii are somewhat irrelevant, give every student a blank sheet of paper, a straightedge, and a compass. (Alternatively, the activity that follows could be completed using a digital geometry tool; one possibility already exists at https://www.desmos.com/geometry/lvbvmaqgm0.) Ask students to create a segment, and then have them construct a circle that is tangent to the segment at its midpoint. When students inevitably ask, “How big should it be?” the proper response is, “Any size you like.” Some students will take this as a challenge to draw the largest circle possible, and others will take it as a challenge to draw the smallest 126

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

circle possible. Both extremes should be welcomed and encouraged because the variety of sizes will help to prove the point that the lengths of the radii are unimportant. Have students do a gallery walk to compare their drawings with one another. Be sure that all students see at least one example of three types: (1) drawings where the inner circle is extremely small so that the annulus and the outer circle are very nearly equal in area, (2) drawings where the inner circle is fairly large so the annulus is a very thin band between the circles, and (3) drawings similar to the one used in the problem stem so the annulus is neither very thin nor very thick. The point of this comparison is to show that the radii of the circles are not fixed, so one strategy is to set one of the radii to any value we like. This is where wishful thinking comes in: If the radii can be any size, is there a particular length that would be advantageous? Indeed! If the radius of the inner circle were 0 cm— that is, if the inner circle were degenerate—then the radius of the outer circle would be 50 cm, and the area of the annulus could be computed easily: π × 502 − π × 02 = 2,500π square units. Of course, choosing the inner radius to be 0 cm would be the most convenient, but wishful thinking allows for any choice of radius length. It would be just as appropriate to choose 120 cm as the inner radius; then the outer radius would be 1202 + 502 = 130 cm, and the area of the annulus would be π × 1302 − π × 1202 = 2,500π square units.

Answer and Solution 2,500π square units. The answer can be found algebraically by assuming that the radius of the outer circle is R and the radius of the inner circle is r. The area of the annulus is πR2 − πr2 = π(R2 − r2). The value of the expression within the parentheses may not be obvious, but the difference of two squares might be a clue. A triangle can be formed using half the chord and the two radii, as shown.

Applying the Pythagorean theorem then gives— R2 = r2 + 502 R2 − r2 = 2,500 Therefore, the area of the annulus is π(R2 − r2) = 2,500π square units.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

127

PROBLEM 74. In the Distance Common Core State Standards for Mathematics: 8.G.B.7 • HSG.C.A.2 • MP.4

In the Classroom The classroom discussion around this problem can focus on two aspects: (1) Emphasize the mathematical modeling that occurs when the situation is represented by a two-dimensional figure; (2) work with students to find a simplified rule that can be used to quickly estimate the distance to the horizon. Begin by having students speculate as to the distance to the horizon from 100 feet above sea level. Many students may not have a valid frame of reference for this, especially for students in landlocked regions; in addition, estimating height above the ground is not a typical skill of most students. Provide a local reference so they can think about the situation. For instance, ask them to think about how far they could see to the horizon if they were standing on the roof of the school. Be prepared for wildly different answers, from as little as several feet to as much as hundreds of miles. Refocus students to consider the distance to the horizon, as their estimates often focus on how far they can see in general.

Once students have a reasonable understanding of the situation, they can construct a model to represent the situation. As Einstein (may have) said, “Everything should be as simple as possible, but no simpler.” Using Occam’s razor, this problem can be represented in two dimensions with a circle for the earth and a tangent for the line of sight to the horizon. (See the image in the solution below.) The important geometric aspect is that the radius of the circle meets the point of tangency at a right angle, thereby creating a right triangle. Once the model is formed, many students will realize that the Pythagorean theorem can be applied. The search for a formula that allows for a reasonable approximation of the distance to the horizon requires students to consider the relationship between algebraic equations and the real-world situation that they represent. If the height above the Earth is h feet, the distance to the horizon is d miles, and the radius of the Earth is assumed to be 4,000 miles, then the situation can be modeled as follows: 128

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Applying the Pythagorean theorem then yields— 4,0002 + d2 = (4,000 + h)2 16,000,000 + d2 = 16,000,000 + 8,000h + h2 d2 = 8,000h + h2 Because d is measured in miles but h is measured in feet, the formula needs to be adjusted. First, it should be noted that the value of h2 will be negligible compared to the value of 8,000h, so for the purpose of an estimate, the value of h2 can be ignored. Moreover, because 8,000 ÷ 5,280 ≈ 1.5, the estimation formula can be rewritten as follows: d = 1.5h



Once a formula for an estimate is available, a follow-up question to ask is “To see 100 miles to the horizon, how high above the ocean would you need to stand?” [about 6,600 feet, or 1.25 miles]

Answer and Solution Approximately 12.6 miles. From any given point on the Earth, the radius is between 3,950 and 3,963 miles; the radius is greater at the equator than at the poles. For the purpose of this estimation, it won’t make much difference which value we use, so let’s assume 3,956 miles as the radius for the location of the lighthouse. The figure below shows the situation. The lighthouse projects 100 feet above the surface of the Earth, and a tangent line represents your line of sight to the horizon. Because tangent lines are perpendicular to the radius at the point of tangency, a right triangle is formed. The longer leg of the triangle is 3,956 miles. The hypotenuse is the distance from the center of the Earth to the lantern 100 = 3956.02 miles. room of the lighthouse, which is 3,956 + 5,280

At this point, you may be thinking that those numbers are so close, it won’t make any difference. But it does! The distance to the horizon, d, is unknown, and we can use the Pythagorean theorem to find it:

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

129

3,956.022 = 3,9562 + d2 158.2 = d2 d ≈ 12.6 miles

PROBLEM 75. Pipe Dream Common Core State Standards for Mathematics: 8.G.B.7 • HSG.C.A.2 • MP.4

In the Classroom When you present the image from this problem to students and ask, “What do you know?” the inevitable response is “Not much.” Give students some time to think individually, and then have them brainstorm with a partner a list of things they know. With effective questions, students come to realize that they know more than they thought they did. The obvious observations will include the following: n

The pipe is a circle.

n

The length of the meter stick is 100 centimeters.

n

The distance from the midpoint to an end of the meter stick is 50 centimeters.

n

The distance from the midpoint to the pipe is 10 centimeters.

You can then ask more pointed questions, such as “Is there anything you know about the angles?” or “Could you add any lines to the figure to show something you know?” With some prodding, the list of observations will grow considerably: n The distance from the pipe to the midpoint is measured perpendicular to the meter stick.

Consequently, those line segments meet at a right angle. n The distance from the center of the circle to an endpoint of the meter stick is a radius of the circle. n

The distance from the center to the midpoint is 10 centimeters less than the radius, or r − 10.

n

The segment from the center to the midpoint meets the meter stick at a right angle.

n A right triangle is formed by half the meter stick, a radius, and the segment from the center to

the midpoint. It is that last observation—which may take a while to get to—that provides a viable solution path. The discussion is a valuable lesson in perseverance for students, and it’s a reminder that getting messy is often the best first step when attempting a problem.

130

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Answer and Solution 260 cm. If the radius of the pipe is r, then a right triangle can be formed by drawing a radius from the center of the circle to one edge of the meter stick and a line segment from the center to the midpoint of the meter stick. The hypotenuse of that right triangle will be r, and the legs are 50 and r − 10.

Applying the Pythagorean theorem then yields— r2 = (r − 10)2 + 502 r2 = r2 − 20r + 100 + 2,500 r2 = r2 − 20r + 2,600 20r = 2,600 r = 130 Since the radius of the pipe is 130 centimeters, the diameter of the pipe is 260 centimeters. Now that we know the answer, it makes perfect sense. The three sides of the right triangle are 50, 120, and 130, which is a multiple of the familiar 5-12-13 Pythagorean triple.

PROBLEM 76. Around the World Common Core State Standards for Mathematics: 7.G.B.4 • MP.4

In the Classroom For classroom demonstration purposes, bring the following objects to class: n

Two circular objects of reasonably different size, for instance, a frisbee and a hula hoop

n

A piece of string or rope, approximately 20 feet long

n

A ruler Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

131

As an analog to the problem about the wire around the Equator, place one of the circular objects on the ground, and carefully lay out the string so that it is 12 inches away from the object at all points. (Even a small object will require an area that is several feet in diameter, so desks may need to be moved, or the activity could be completed in the hallway.) Measure the length of the string. Then, determine the circumference of the object by wrapping string around the object, and measure the length of the string. Record both measurements and the difference between them in a table. Object

Circumference of Object

Length of String One Foot Away

Difference

frisbee

38 inches

110 inches

72 inches

hula hoop

Repeat the process for the second object and others, as necessary. The difference between the two measurements should be similar in all cases. Any discrepancy is due to imprecision, both in terms of measurement as well as how accurately the circle 1 foot away from the circumference was created. Note that it may take more than two objects to convince students that the difference will always be the same—a little more than 75, or 24π, inches—regardless of the size of the objects. The important takeaway is that the difference between the circumference of the object and the circle 12 inches away is constant, regardless of the circumference of the object. The difference is based entirely on the distance from the string to the object.

Answer and Solution 200π meters. The radius of the Earth at the Equator is approximately 6,000,000 meters. (As we’ll see in a minute, the exact radius doesn’t really matter.) Because the piece of wire will be suspended 100 meters above the Earth’s surface, its radius will be 6,000,100 meters. Consequently, the circumference of the Earth is 6,000,000 × 2π = 12,000,000π, and the length of the wire is 6,000,100 × 2π = 12,000,200π. Therefore, the wire is only 200π meters greater than the circumference of the Earth!

Interestingly, a wire suspended 100 meters above the surface of any sphere would be 200π meters greater than the circumference of the sphere, regardless of the size of the sphere. To see that this is 132

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

true, assume the radius of the sphere is R meters, so its circumference is 2Rπ. Then the length of the wire is R + 100, and its circumference is 2(R + 100)π = 2Rπ + 200π, which is 200π meters greater than the circumference of the sphere.

PROBLEM 77. The Setting Sun Common Core State Standards for Mathematics: 7.G.B.4 • HSG.C.A.2 • MP.2

In the Classroom An incredibly powerful problem-solving strategy is wishful thinking. In this problem, the radii of the small semicircle and the large semicircle are unknown. Wouldn’t it be nice if both radii were known? By employing wishful thinking, it’s possible to make this a much simpler problem. If the radius of the small semicircle were 0 units, then the radius of the large semicircle would be 50 units, and the shaded region would be the area of the entire large semicircle with nothing removed. The area would be 12 × π × 502 = 1,250π square units. The question is whether you are allowed to do that. Can you just assume that the chord is a radius? Indeed, you can. As shown in the solution below, the area is 1,250 square units regardless of the lengths of the radii. The variables R and r are used to represent the large and small radii, respectively, but their exact lengths are irrelevant to the solution.

Answer and Solution 1,250π square centimeters. The radius of the large semicircle is R, and the radius of the small semicircle is r. But more important, the distance between the two parallel lines is r at all points. This allows us to construct a right triangle with hypotenuse R and legs of length r and 50 as shown below:

Applying the Pythagorean theorem yields the following: R2 = r2 + 502 R2 − r2 = 2,500 Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

133

The area of the shaded region is the difference between the area of the large semicircle minus the area of the small semicircle, which is 1  πR2 − 1  πr2 = 1  π(R2 − r2). From above, we know the 2 2 2 value of the expression within the parentheses, so the area of the shaded region is 1  π(R2 − r2) = 2 1  π(2,500) = 1,250π. 2

PROBLEM 78. Growing Pattern Common Core State Standards for Mathematics: 5.OA.A.2 • 7.EE.A.2 • HSA.SSE.A.1

In the Classroom This problem could, and probably should, be introduced by showing Stage 3 only. Students could be asked to consider how to determine the total number of squares in the figure without counting them one by one. To orchestrate a classroom discussion, ask students to think about how they would determine the number of squares and to place a closed fist in front of their chest when they’ve determined at least one method. Then, ask students to raise a thumb on their fist if they are ready to share their thoughts with a partner. When students are ready, ask them to discuss their method with others. After sharing with a partner for a minute or two, convene the whole class and ask, “Is anyone willing to share a method for counting the squares that was suggested by their partner?” (This equity tactic helps to prevent high-achieving students from dominating the discussion.)

Students will identify a variety of ways to count the squares, and three possibilities are shown above. Some students might notice a 3 × 3 grid with three squares above and three squares to the right, as in figure A; this leads to the expression (3 × 3) + 3 + 3. Other students might notice a 4 × 3 grid with three squares to the right, as in figure B; this leads to the expression 4 × 3 + 3. (Incidentally, some might instead notice a 3 × 4 horizontal rectangle with three squares above, which would lead to the expression 3 × 4 + 3.) Still other students might notice that the three squares along the right, as shown in figure C, could be moved above or below the 4 × 3 rectangle, thus creating a 5 × 3 rectangle; this leads to the expression 5 × 3. Although not represented above, another method often suggested by students is to recognize that the figure is a 4 × 4 grid with one 134

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

square removed from the upper right corner; this leads to the expression 4 × 4 − 1. All of these methods are valid, and it’s empowering for students to share their methods with one another. The role of the teacher during this discussion should be to capture student thinking and provide a correct mathematical expression to accompany each student’s method. While it’s possible for students to generate their own expressions, the teacher can model the use of appropriate notation. After several methods for counting the squares in Stage 3 have been shared, the entire problem stem —without the associated question about Stage 100—can be presented to the class. The teacher could ask, “What do you notice? What mathematical questions do you have? What mathematical questions could you answer?” Some questions that typically arise— n

What does Stage 0 look like?

n

Does the pattern continue forever?

n

What does Stage 4 look like? What does Stage 50 look like? What does Stage n look like?

n

How many squares are in Stages 1, 2, and 3?

n

How many squares are added at each stage?

From there, a discussion can ensue about which of the methods for counting the squares in Stage 3 would also work to count the squares in Stage 1 and Stage 2. Many of the methods will work, and most can be generalized. For instance, the method of moving three squares to form a 5 × 3 rectangle could lead to the generalization n × (n + 2) for the number of squares in Stage n. (The figure used in Stage 3 comes from Visual Mathematics, Course 1, Lesson 5 (Cooper Foreman and Bennett 1995, p. 51). The idea for extending it to a growing pattern comes from Nicole Rigelman [personal communication, January 28, 2020]).

Answer and Solution 10,200 squares. The shading of the figure below indicates one possible way to generalize.

Notice the pattern: n

In Stage 1, there is a 1 × 1 grid of squares with one square above and one square to the right.

n

In Stage 2, there is a 2 × 2 grid of squares with two squares above and two squares to the right.

n In Stage 3, there is a 3 × 3 grid of squares with three squares above and three squares to the right. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

135

This generalizes to the following: n In Stage n, there is a n × n grid of squares with n squares above and n squares to the right, and

the total number of squares in Stage n is n × n + n + n = n2 + 2n. In Stage 100, the total number of squares will be 1002 + 2(100) = 10,200.

PROBLEM 79. Survivor, Part 1 Common Core State Standards for Mathematics: 1.OA.C.6 • MP.7

In the Classroom This problem is a variation of the math game Nim that was used on the TV show Survivor Thailand in Fall 2002. In that game, only 21 flags were used, but the puzzle was otherwise the same. A coin toss determined which team would go first, and the team going first could have ensured victory by using the correct strategy. Sadly, the team either didn’t know the strategy or applied it incorrectly, and they lost. In class, this game can be played easily with any type of manipulative: coins, tokens, buttons, even strips of paper. As with the Survivor version, it might make sense for students to play the game several times using fewer than 100 objects; 21 is a good number. The first time they play, students seemingly choose entirely at random. By the second or third time, they’ve usually deduced an end-game strategy, realizing that leaving three objects for their opponent will result in a win. It may take four, five, or more games before students realize that the game can be completely solved: The strategy is to leave the opponent with a multiple of three flags on every turn. To ensure that students don’t tire of the game before analyzing the strategy, have them play against different opponents to keep engagement high. For young kids, this game provides an opportunity to practice basic subtraction facts. The game can be modified so a different number of objects is taken each time; this can be useful so that students get practice with a variety of number facts. Further, the game can become more educationally useful if equations are recorded while playing, such as 11 − 3 = 8. For very young students, a game with only 10 flags could be sufficient to help with fluency. This game, and all Nim variants, highlight backward induction, the process of working backwards from the end of a problem to determine the optimal move at each point in time. In sequential games, the process involves two players and leads to an optimal strategy. On the other hand, game theory studies more complex interactions among many players, but always with the intent of anticipating what each player will do on a given turn. Consequently, Nim, Kayles, Oddly, Wythoff’s Game, and many other sequential take-away games can lead to excellent math discussions and provide an introduction to game theory. 136

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Answer and Solution Remove one flag on your first turn to guarantee a win. This game is one of many variations of Nim, a mathematical game in which players take turns removing objects from several distinct piles. In this version, there is just one pile with 100 objects. To determine a winning strategy, consider a simpler problem. If, after some number of turns, you reduced the number of flags to just one, your opponent would win; they would take the remaining flag. Similarly, if you reduced the number of flags to two, your opponent would also win; per the rules of the game, they could take both of the remaining flags. However, if you reduced the number of flags to three, you could win: If your opponent then took one, you would subsequently take the last two, or if your opponent took two, you would win by taking the last one. This logic can then be repeated. If you left your opponent with either four or five flags, they could win by taking either one or two flags to reduce the number to three. Then, according to the logic above, they’d win. But if you left your opponent with six flags, you would win; your opponent would take either one or two, and then you’d take either two or one to reduce the pile to three. This logic can be repeated, and it can be seen that if you can leave your opponent with any multiple of three, then you can always reduce the pile to a lower multiple of three on your next turn. This is possible because you can remove either one or two flags on each turn, so every pair of adjacent turns—one by your opponent, one by you—can reduce the total number by 1 + 2 = 3 flags. Eventually, you’ll leave just three flags, and win. The multiple of three closest to 100 is 99. Therefore, you should take just one flag on your first turn to guarantee a win. On the other hand, if your opponent goes first, you’ll have to hope that they either don’t know the optimal strategy or just make a mistake. Otherwise, they’ll win.

PROBLEM 80. Survivor, Part 2 Common Core State Standards for Mathematics: 2.OA.A.1 • 4.OA.C.5 • MP.3

In the Classroom Nothing substitutes for experience, so allow students to play several rounds of this game. (In fact, you could use this game to simulate the single-elimination tournament described in problem 94, One and Done.) Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

137

Students will start to recognize an end-game strategy, and a class discussion about this can be valuable. Ask, “When you get near the end of the game, what are some good numbers at which to end your turn? What are some bad numbers at which to end your turn?” Students will offer that 2 is a good number because your opponent can only then take 1, leaving 1, a bad number. The follow-up question, then, is critical: “From what numbers can you get to 2?” Allow students to discuss in pairs before sharing with the whole group. Some students will realize that 7 is a good number, because your opponent cannot take 8 coins; if he takes 1 coin, you can take 4 coins to land at 2, and if he takes 4 coins, you can take 1 to also land at 2. During this discussion, some students may also suggest that 5 is a good number, because it’s then possible to get to 0 on your next turn regardless of your opponent’s move. This is all good insight, and you can model an organized approach to analyzing the entire game. On the board or a piece of chart paper, list the numbers from 1 to 100. Circle the numbers that students have identified as “safe,” and cross out those numbers identified as “unsafe.” 1  2  3 4  5  6  7  8 9 10 11  12  13 . . . 100 After some initial work together, students can be left to complete the analysis in small groups.

Answer and Solution Four coins. In this game, there are safe and unsafe numbers. For example, 1 is unsafe because if you leave just 1 coin for your opponent, you’ll lose. On the other hand, 2 is safe, because if you leave 2 coins for your opponent, then your opponent must take 1 coin, leaving just 1 coin and resulting in a win for you. A complete analysis yields the following safe numbers: 0, 2, 5, 7, 12, 14, 17, 19, 24, 26, 29, 31, 36, 38, 41, 43, 48, 50, 53, 55, 60, 62, 65,67, 72, 74, 77, 79, 84, 86, 89, 91, 96, 98. Consequently, the first player should take 4 coins on the first turn to leave 96 coins, which will then guarantee a win (assuming she continues to play with optimal strategy). Of note, the sum of two consecutive moves could be 1 + 1 = 2, 1 + 4 = 5, 1 + 8 = 9, 4 + 4 = 8, 4 + 8 = 12, or 8 + 8 = 16. The list of safe numbers above makes it obvious that if you leave one of the safe numbers from that list, then no matter what your opponent takes, you can get to another safe number on your next turn. For instance, if the game proceeds to the point where you leave your opponent with 55 coins (a safe number from the list), then your opponent has three options: (1) Take 1 coin; then you can take 1 coin to leave 53 coins (another safe number); (2) Take 4 coins; then you can take 8 coins to leave 43 coins (yet another safe number); or (3) Take 8 coins; then you can take 4 coins to again leave 43 coins.

PROBLEM 81. Survivor, Part 3 Common Core State Standards for Mathematics: 4.OA.C.5 • HSF.IF.A.3 • MP.3 138

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

In the Classroom The original version of this problem, known historically as the Josephus problem after the Roman historian Flavius Josephus, uses a more gruesome context involving the execution of soldiers. Information about it can be found at http://mathworld.wolfram.com/JosephusProblem.html. In general, the Josephus problem asks which person will remain when the initial circle starts with n people and every kth person is removed. Presented here is the specific case for n = 100 and k = 2. A good classroom activity is determining the general rule for which player remains when n people start in the circle and every second person is removed. A simulation can be conducted in class for small numbers of people. To gather some initial data quickly, create small groups with a different number of students in each. Groups with 3, 4, 5, and 6 work well to start; with each group, include one additional student who will be responsible for removing students from the circle. The data should be captured in a whole-class table. The simulation can then be completed with larger groups, though at some point, it will be far more efficient and less disruptive for students to perform simulations at their desks. Small pieces of paper with the numbers 1 through n can be used. As more simulations are completed, data can be added to the whole-class chart. Students will start to notice patterns when n reaches the teens. n

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

Survivor

1

1

1

1

3

5

7

1

3

5

7

9

11

13

15

1

The survivor for 2n is always 1, and a pattern consisting of the odd numbers proceeds until the next power of 2 is reached. That insight will lead to an answer, but a valuable follow-up exercise is asking students to write a formula that gives the survivor S(n) when the circle has n people to start.

Answer and Solution Player 73. Number the players who start the game from 1 to 100. During the first round, every person with an even number leaves. At this point, after the first round, you may note that the players remaining are those whose numbers leave remainder 1 when divided by 2, that is, the odd numbers from 1 to 99; after the second round, the following players remain: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69, 73, 77, 81, 85, 89, 93, 97 After the second round, the players remaining are those whose numbers leave remainder 1 when divided by 4. After the third round, the following players remain: 1, 9, 17, 25, 33, 41, 49, 57, 65, 73, 81, 89, 97 Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

139

Those players are the ones whose numbers leave remainder 1 when divided by 8. After the fourth, fifth, and sixth rounds, respectively, the following players remain: 9, 25, 41, 57, 73, 89 → 9, 41, 73 → 73

PROBLEM 82. Complete the Circuit Common Core State Standards for Mathematics: 4.OA.C.5 • MP.3 • MP.8

In the Classroom As with problem 99, Squares on a Grid, this problem can be presented as a group problem-solving activity. On a whiteboard or piece of chart paper, write the numbers 1 to 100, or simply display a laminated hundreds board. This will be used as a class recording sheet to indicate the numbers for which a circuit has been found. Display the image of the grid at the front of the class, and then read the problem stem without the problem. That is, explain possible paths and how a complete circuit is formed, but don’t ask the question about which ones are possible. Instead, show one possible circuit, and then have students attempt to find a different circuit on their own. After all students have found one, then let them work in pairs or small groups for several minutes to find others. Circulate as students work. As they proceed, some students will start to see the alternating pattern of numbers that are possible. Allow them to discuss in their small groups without yet sharing with the class. When all groups have made progress, reconvene the class and call on several students to tell you one of the numbers for which they have found a circuit. As verification, ask students to share the circuit that led to each number. (This is a good opportunity to promote student agency. As you circulate, note reluctant students who have found a circuit, and call on them first during the wholeclass discussion.) Periodically pause to ask if students can identify a pattern. Once they do, deliver the payoff pitch by asking them to justify their conjecture. One possible justification is given in the solution below.

Answer and Solution 3,025. If the squares of the grid are colored like a chessboard, analyzing the situation becomes much easier. From the cell numbered 1, which is white, it’s only possible to move to cell 2 or cell 11, both of which are black. From there, it’s only possible to move to a white square. And so on, alternating

140

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

white, black, white, black, . . ., until a circuit is complete. Consequently, any complete circuit must end on a black square.

The question is, Is it possible to land on any of the black squares? In fact, it is. The figure below (left) shows that it’s possible to create a circuit that ends at 91. The last 13 cells in that circuit are 64, 74, 84, 94, 93, 83, 73, 72, 71, 81, 82, 92, 91. The figure below (right) shows a possible modification so that the last 13 cells in the circuit instead are 64, 74, 73, 72, 71, 81, 91, 92, 82, 83, 84, 94, 93, with the circuit ending at 93.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

141

With similar modifications, it would be possible to end on any odd number in the bottom row; moreover, it would be possible to land on any black square in the entire grid. For example, the following circuit ends at 37:

What is the sum of the numbers on all of the black squares? Note that there are five black squares in each row and in each column. Therefore, the sum of all black squares is the following: 5 × (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) + 5 × (10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100) = 3,025.

PROBLEM 83. Numbering the Vertices Common Core State Standards for Mathematics: 4.OA.C.5 • MP.7 • MP.8

In the Classroom This problem is a modified version of a problem in Tanton’s (2013) Without Words (p. 67). In that book, the author presents the puzzle without any words—just an image of seven dots connected so that each vertex has a different degree. The problem statement above explains much of the vocabulary needed to communicate about this problem and its solution, but the problem could also be presented as just the two visuals without any explanation. Students would need to make observations and look for patterns. Most students will see the relationship between the numbers and the degree of each vertex rather quickly. To construct analogous figures with different numbers of vertices will require that students get a little “dirty.” Zucker (2017) says that when it comes to problem solving, students need to get dirty: 142

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Patience alone is not enough: you have to do something while you’re being patient! [. . .] Call it “Getting Dirty.” Which is to say, you can’t expect to fix your plumbing problem if you insist on your hands staying clean the whole time. Try special cases: doodle, try some examples, try some different (easier) versions of the problem, try some numbers, guess and check (2017, p. 2).

The mathematics within the problem is completely accessible to most students, but the context from graph theory may not be. Consequently, students will need to investigate before they start to see any patterns. Students can work in pairs or small groups to explore graphs with 4, 5, 6, or 7 vertices. Such an investigation may require drawing, erasing, and starting over. When students find that they are unable to create graphs with 5 or 6 vertices, pose the question, “If you’re sure that it’s not possible, can you explain why not?” As a follow-up, you can also ask, “For what other numbers of vertices would it be impossible to create a graph? How do you know?” Because this problem will not be solved quickly by most students, it’s important to continually prompt further investigation with questions rather than reducing the cognitive demand by offering statements that lead to a particular solution. Perseverance can only be developed and demonstrated if students are given ample opportunity to struggle.

Answer and Solution Yes, it’s possible. The problem statement indicates that it’s possible to create graphs with three and eight vertices such that each vertex has a different degree, and a little playing will show that it’s also possible for graphs with 4 and 7 degrees. On the other hand, it’s not possible to create similar graphs with one, two, five, or six vertices. It seems, then, that the pattern of possible graphs is with 3, 4, 7, 8, 11, 12, 15, 16, . . ., vertices, but is there any reason to believe that this pattern actually holds? Each edge in a graph increases the total degrees of all vertices by 2, because each of the two endpoints of an edge occurs at a vertex. Consequently, the sum of the degrees of all vertices in a graph must be even, so it’s only possible to create a graph with n vertices that have degrees 1 though n if the value of 1 + 2 + 3 + ⋯ + n is even. Consider the case for n = 2. The sum is 1 + 2 = 3, but it’s not possible to create a graph where the sum of all degrees is 3. If there are two edges, the sum of all edges would be four, and if there were just one edge, the sum of all edges would be two. For what values of n will 1 + 2 + 3 + ⋯ + n be even? Consider the formula for the sum of the first n positive integers. From problem 21, Gauss and Check, that formula is as follows: 1 + 2 + 3 + ⋯ + n = (n)(n2 + 1) Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

143

Because the values in the table give us reason to believe the pattern repeats every fourth number, examine the results for four consecutive numbers of the form 4k, 4k + 1, 4k + 2, and 4k + 3. (4k)(4k + 1) 16k 2 + 4k = 2 = 8k2 + 2k, which is even. 2 (4k + 1)(4k + 2) 16k 2 + 12k + 2 = n If n = 4k + 1, then = 8k2 + 6k + 1, which is odd. 2 2 2 (4k + 2)(4k + 3) 16k + 20k + 6 = n If n = 4k + 2, then = 8k2 + 10k + 3, which is odd. 2 2 (4k + 3)(4k) 16k 2 + 12k = n If n = 4k + 3, then = 8k2 + 6k, which is even. 2 2 n If n = 4k, then

This confirms that the pattern will, indeed, repeat in groups of 4. Moreover, since 100 = 4 × 25, it is of the form n = 4k. Consequently, the sum will be even, implying that it’s possible to create a graph with 100 vertices so that the vertices have degrees 1 through 100.

PROBLEM 84. Shake It Off Common Core State Standards for Mathematics: 4.OA.C.5 • MP.1 • MP.4

In the Classroom As noted by Naylor (n.d.), this “problem has an interesting context with the Supreme Court. This lesson works well if used near the first Monday in October, because that is the date that the Supreme Court convenes each year. [. . .] One of [the Supreme Court’s] traditions is that every justice shakes hands with each of the other justices each time they gather for a meeting. Chief Justice Melville W. Fuller (1888−1910) started this custom, saying that it shows ‘that the harmony of aims, if not views, is the court’s guiding principle.’” The Supreme Court has nine justices, so students can act out the scenario in which nine people shake hands with one another. To fully understand the problem and to be able to predict how many handshakes occur with 100 people, students can look for a pattern in the number of handshakes that occur with different numbers of people. An effective way to fully explore the mathematics behind this situation in the classroom appears on the NCTM Illuminations site, https://www.nctm.org/Classroom-Resources/ Illuminations/Lessons/Supreme-Court-Handshake/.

Answer and Solution 4,950 handshakes. Pick one person at random; she will shake 99 hands. The next person will only shake 98 hands; he doesn’t need to shake the hand of the first person. The next person will shake 97 hands; the person after that will shake 96 hands; and so on. The total number of handshakes will be 99 + 98 + 97 + ⋯ + 3 + 2 + 1 = 4,950. 144

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

In general, the sum of 1 + 2 + 3 + ⋯ + n will consist of 2n pairs of numbers that add up to n + 1. That is, adding the first and last terms gives n + 1; adding the second and next-to-last terms gives (n − 1) + 2 = n + 1; and so on. Thus the sum is 12(n)(n + 1). For n = 99, the sum is 12 (99)(99 + 1) = 4,950.

PROBLEM 85. Scope It Out Common Core State Standards for Mathematics: 4.OA.C.5 • 5.NF.A.1 • HSS.CP.B.9 • MP.8

In the Classroom One effective solution strategy is to solve a simpler problem and look for a pattern. Typically, students quickly realize that the sum of the first n terms is n n+ 1 . Recognizing that the patterns holds is one thing, proving that the formula always works is another. Various proofs provide an opportunity to use multiple representations. When this problem was shared with a middle school math club, one student suggested that the nth term in the series is equal to n n+ 1 − n n− 1 . Rewriting each term in this way yields a telescoping series in which a part of one term cancels with a part of the next term. Hence the name telescoping because the series will collapse on itself. The telescoping series formed is—  1 0  2 1  3 2  99 98  99 0 99  2 − 1  +  3 − 2  +  4 − 3  +  +  100 − 99  = 100 − 1 = 100 Note that the first fraction in each term will cancel with the second fraction of the following term. 99 from the last term. After terms cancel, all that will remain are the 01 from the first term and 100 It’s also possible to represent the series with a visual model. The series can be rewritten as— 1 1 1 1 1 1 1 1 × + × + × ++ × 1 2 2 3 3 4 n n +1 and each product can be represented successively in a square. For instance, the first product is 11 × 12 , which represents one-half of the whole. This can be shown by shading half of an entire square. Similarly, the next product is 12 × 13, which can be shown by shading one-third of the other half. These two steps are shown in the following figure:

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

145

Continuing in this manner, one-fourth of the remaining one-third, then one-fifth of the remaining one-fourth, and so on, can be shaded. After n sections have been shaded, the unshaded portion will be 1n , so the entire shaded portion is 1− 1n = (n −n 1) .

Eventually, nearly the entire square will be shaded, indicating that the limit of the infinite series would be 1. A third, and the most common, proof of the formula involves recognizing that the nth term of the series is (n)(n1 + 1) . You might ask students if they can write this fraction as the difference of two fractions. Some students may realize that (n)(n1 + 1) = 1n − n 1+ 1; if not, you can provide students with the denominators and ask for the numerators. If each term is then rewritten, a different telescoping series is formed. In this series, when each term is rewritten as a difference of fractions, the result is the following:  1 1   1 1  1 1   1 1  1 1  1 1 99  1 − 2  +  2 − 3  +  3 − 4  +  4 − 5  +  +  99 − 100  = 1 − 100 = 100 The − 12 in the first term cancels with the 12 in the second term; the −13 in the second term cancels with the 13 in the third term; and so on. Only the first fraction in the first term and the second fraction in the last term will remain after all the cancellations happen. This idea can be used to show that a general formula holds, too: 146

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

 1 1   1 1  1 1   1 1  n 1  1 − 2  +  2 − 3  +  3 − 4  +  n − n + 1 = 1 − n + 1 = n + 1 An alternative proof uses induction. The conjecture is that the formula is 1 1 1 1 n 1 n 1 + + ++ = 1× 2 2 × 3 3× 4 (n)(n + 1) n + 1 , and the formula clearly holds for n = 1: 1 × 2 = n + 1 = 2 . Then, assuming the formula holds for n = k, can we show that it also holds for n = k + 1? Indeed, we can. 1 1 1 1 1 k 1 + + ++ + = + (k)(k + 1) (k + 1)(k + 2) k + 1 (k + 1)(k + 2) 1× 2 2 × 3 3× 4 =

k × (k + 2) + 1 (k + 1)(k + 2)

=

k 2 + 2k + 1 (k + 1)(k + 2)

=

(k + 1)(k + 1) (k + 1)(k + 2)

=

k +1 k+2

Answer and Solution 99 . Look for a pattern. 100 1 1 = 1× 2 2 1 1 2 + = 1× 2 2 × 3 3 1 1 1 3 + + = 1× 2 2 × 3 3× 4 4 1 1 1 1 4 + + + = 1× 2 2 × 3 3× 4 4 × 5 5 In general, if the denominator of the nth term is (n)(n + 1), then the sum of the first n terms is n n+ 1. 99 . Therefore, the 100th term will be 100

PROBLEM 86. Erase and Replace Common Core State Standards for Mathematics: HSF.IF.A.2 • HSF.BF.A.1.A • HSS.CP.B.9 • MP.3

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

147

In the Classroom It doesn’t take long for the numbers in this problem to get fairly large. That’s a clue to students that they ought to try a simpler problem and look for patterns. If you suspect that students will get frustrated using the numbers 1 to 100, suggest that they try the game with a smaller set. Several students playing the game with the same set of numbers will soon realize that they always get the same single-number result. They may even notice that the result is always 1 less than a factorial. They may, however, have difficulty with two aspects: 1. Convincing themselves that the result will always be the same, no matter the order in which numbers are erased and replaced 2. Proving that the generalization is true in all cases, that is, justifying the conjecture that the single-number result will always be 1 less than a factorial Proving that the result always holds is the less difficult of the two. This is simply a matter of realizing that the given operation is associative. Because the process involves addition and multiplication, both of which are associative, it seems likely that the operation described in this problem would be associative, too. To see that this is true, assume that three numbers, x, y, and z, remain on the board. For convenience, define the operation a @ b = a + b + ab. If x and y are erased first, the following would be the result: (x @ y) @ z = (x + y + xy) @ z = x + y + xy + z + xz + yz + xyz Alternatively, if y and z were erased first, the result would be exactly the same: x @ (y @ z) = x @ y + z + yz = x + y + z + yz + xy + xz + xyz This proves that the result is associative, so the order in which the numbers are erased is irrelevant. An outline of a proof showing that the result will always be 1 less than a factorial is given in the solution below.

Answer and Solution 101! − 1. Let’s solve a simpler problem first. What if only the numbers 1 and 2 were written on the board? When erased, they would be replaced with 1 + 2 + 1 × 2 = 5. Now, what if the numbers 1, 2, and 3 were on the board? Well, 1 and 2 could be erased to give 5; with 3 and 5 now left on the board, they would be erased to give 3 + 5 + 3 × 5 = 23. Continuing in this manner, a pattern quickly emerges: If the integers from 1 to n are on the board, the last remaining number will equal (n + 1)! − 1. For the integers 1 to 100, the only possible value of the remaining number is 101! − 1. 148

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

How the pattern arises may not be totally clear. Consider the general case with two numbers a and b. When erased, they are replaced by— a + b + ab, which can be rewritten as a + b + ab + 1 − 1. Adding and subtracting 1 may seem arbitrary, but it’s another example of wishful thinking: adding 1 yields an expression that can be factored nicely and subtracting 1 is just keeping things in balance. The expression above can be rewritten as— a + ab + b + 1 − 1 = a × (1 + b) + 1 × (b + 1) − 1 = (a + 1)(b + 1) − 1 This expression implies that we will continually get 1 less than a factorial. If a = 1 and b = 2, then the expression becomes (1 + 1)(2 + 1) − 1 = 3! − 1 = 5. Now that 5 is a member of the collection, let a = 5 and b = 3, and the result is (5 + 1)(3 + 1) − 1 = 3! × 4 − 1 = 4! − 1 = 23. We could continue in this manner ad infinitum, and the result will always be (n + 1)! − 1 for the integers from 1 to n.

PROBLEM 87. Replace the Hypotenuse Common Core State Standards for Mathematics: 8.G.B.7 • HSG.SRT.B.4 • HSF.BF.A.1 • MP.7

In the Classroom The benefit of using this problem in a middle school classroom is giving students practice with the Pythagorean theorem. But instead of completing a worksheet of exercises with naked numbers, applying the Pythagorean theorem in this problem comes with the higher purpose of attempting to identify a pattern and solve the problem. As an extension, ask students to find the two values of n such that if the whiteboard started with the numbers 1 through n, the lone number remaining after erasures would be an integer. In the trivial case, the result would be an integer if n = 1. The only other value that yields an integer is = 70. This was first proved by English mathematician George Neville n = 24, because 24(25)(49) 6 Watson in 1918. Although less complicated proofs have been discovered since, none would be considered elementary.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

149

Answer and Solution 338,350 . Begin at the beginning and see what happens. Instead of using 1 to 100, consider the result when using the numbers 1 through n: 2 5

3

4

5

6

7

8

9

14

30

55

91

140

204

285

If the sequence of numbers underneath the square root signs looks familiar, that’s because you’ve seen it before—problem 24, Square Deal. It’s not surprising that these two problems are related, since problem 24 involves a sum of squares, and this problem involves the Pythagorean theorem, which is based on squares. The sequence 1, 5, 14, 30, 55, . . ., is known as the square pyramidal numbers, and a longer list can be found online at https://oeis.org/A000330. The nth square pyramidal number is given by the formula (n)(n + 1)(2n + 1) ÷ 6, so the 100th square pyramidal number is (100)(101)(201) ÷ 6 = 338,350. Hence, the eventual result when two numbers are continually removed and replaced by the hypotenuse is the square root of that value, or 338,350.

PROBLEM 88. An Odd Game Common Core State Standards for Mathematics: 4.OA.C.5 • MP.3 • MP.7

In the Classroom The hallmark of a good problem is that it naturally leads to other problems. The analysis required to determine that it’s impossible to win when going first and the result must be an odd number is a substantial exercise in number theory. But beyond the initial problem, the following extensions are worth exploration:

150

n

Can you win if the game is played with numbers other than 1 to 100? If so, for what numbers?

n

Is it possible to win if you go first and win when the result is even?

n

I f the game is extended to three players—the first person wins if the result is a multiple of 3, the second player wins if the result is congruent to 1 modulo 3, and the third player wins if the result is congruent to 2 modulo 3—is there an optimal strategy?

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Answer and Solution Forfeit. Unfortunately, there is no possible way to obtain an odd result. Consider the case where only addition signs (+) are added between numbers. The result is 1 + 2 + 3 + ⋯ + 100 = 5,050 (see problem 21, Gauss and Check). Now change one addition symbol to a subtraction symbol to see what happens. If the minus symbol appears before the integer k, then the final sum will be 5,050 − 2k. This obviously changes the value of the result, but it does not change the parity of the result. That is, 5,050 is even, and because the amount subtracted is an even number—clearly, 2n is even because it’s a multiple of 2—then the new result will also be even. The same logic applies whether 1, 5, 50, or all 99 operators are changed; the result may eventually become negative, but it will never be odd. Sorry. The only two ways to win are to change the rules so if the result is even, you win, or to play with numbers other than 1 to 100.

PROBLEM 89. Pick ‘n’ Add Common Core State Standards for Mathematics: 3.NBT.A.2 • MP.8

In the Classroom Students will enjoy this game a thousand times more than completing a worksheet of addition problems, yet it provides the practice they need to develop computational fluency. Playing just two games is the equivalent of completing 10 to 20 exercises involving one- and two-digit addition. To provide additional practice, adjust the game to use different numbers. For instance, students could choose numbers between 50 and 100 to start, and the first player to reach a number greater than 1,000 wins. Similarly, adjust the game to use a different operation. For instance, game play will be similar but with different results and strategy if multiplication is used instead of addition. Because the numbers grow rapidly with multiplication, choose initial numbers from 1 to 10 but play till one player reaches 10,000 or more.

Answer and Solution 8. If one player chooses 8 and the other chooses a different number, the player choosing 8 will always win. The table below shows the winner depending on which numbers each player (A and B) chooses. The only number choice that yields a win no matter what the other player chooses is 8—unless the other player also chooses 8, in which case both players will need to select again. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

151

Number Choice for A 1

Number Choice for B

1

2

3

4

5

6

7

8

9

10

B

A

A

B

B

B

A

A

A

A

B

B

B

A

A

A

A

B

B

A

A

A

A

A

B

A

A

A

A

A

A

A

A

A

B

A

A

B

B

A

B

B

B

B

2

A

3

B

B

4

B

A

A

5

A

A

A

A

6

A

A

B

B

B

7

A

B

B

B

B

B

8

B

B

B

B

B

B

B

9

B

B

B

B

B

A

A

A

10

B

B

B

B

A

A

A

A

B A

PROBLEM 90. Product of Piles Common Core State Standards for Mathematics: 4.OA.C.5 • 4.NBT.B.5 • MP.8

In the Classroom Exploring the problem with 100 tokens can be overwhelming, so looking for a pattern by investigating smaller piles would be a good strategy. For classrooms with younger students, differentiation can happen by allowing students to choose the pile size that they’d like to investigate: 4, 6, 10, or 20. With just four tokens, the process takes only three steps, and the only possible multiplications are 1 × 2, 1 × 3, and 2 × 2. This can be a confidence builder for struggling students. With 6 tokens, the products remain relatively small, but the process requires five steps, so there is more record keeping and opportunities for error. With 10 tokens, students will never be confronted with products larger than 25. With 20 tokens, students can challenge themselves in the first few steps by dividing the piles into amounts that will yield products with which they are not yet familiar, such as 17 × 3 followed by 13 × 4; in addition, the process will require 19 steps, so students who attempt this scenario should have guidance on how to keep track of the results. Giving students a choice of numbers is important to ensure that they are engaged and attempting problems at a developmentally appropriate level. In particular, Drake (2017) suggested that there are four considerations for how to choose numbers: 152

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

1. Consider the numbers in relation to your students on the basis of their current mathematical understandings. 2. C  onsider the numbers in relation to your learning goals and standards since different numbers provide different opportunities. 3. C  onsider the numbers in relation to each other as different numbers will reveal different aspects about our number system. 4. Consider the numbers in relation to the context to ensure that they are reasonable for the situation and match the real world. Providing the opportunity to explore this situation with various numbers allows students to select a problem at an appropriate level of difficulty. Of course, some students may deliberately choose numbers that are below their current level in which case you’ll need to push them to push themselves.

Answer and Solution 4,950—the only possible sum. Beginning with n coins, the sum of the products will be equal to the (n − 1)st triangular number. (Try it yourself with a smaller number of coins to see that this is true.) The formula for the nth triangular number (see problem 21, Gauss and Check) is 1 (n)(n+1), 2 so the 99th triangular number is 1 (99)(100) = 4,950. 2

PROBLEM 91. Magic Rectangles Common Core State Standards for Mathematics: 3.NBT.A.2 • 4.NBT.B.5 • MP.1

In the Classroom One of the benefits of exploring magic rectangles in the elementary grades is that they provide practice with addition as students attempt to find the sums, but they also provide practice with multiplication using an array model to construct the rectangles. For this problem, students are asked to construct rectangles containing 100 squares, but magic rectangles of other sizes could be explored to provide practice with other products.

Answer and Solution Two magic rectangles. 2 × 50 and 10 × 10. The integers 1 to 100 have a sum of 5,050. This sum can be determined by noticing that the numbers can be regrouped as 50 pairs of numbers with a sum of 101 each. (1 + 100) + (2 + 99) + (3 + 98) + ⋯ + (50 + 51) = 5,050 Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

153

In general, the sum of the first n positive integers is given by the following formula: 1 2

S(n) = (n2 + n) There are only four possible sizes for magic rectangles with the numbers 1 to 100. 1. 1 × 100 2. 2 × 50 3. 4 × 25 4. 10 × 10 Because the sum of the integers 1 to 100 is 5,050, it would not be possible to place them in a 4 × 25 array to form a magic rectangle. The sum of each row would have to be 5,050 ÷ 4 = 1,262.5, which is impossible. On the other hand, they could be placed in a 2 × 50 array, where the sum of each row would have to be 5,050 ÷ 2 = 2,525 and the sum of each column would have to be 5,050 ÷ 50 = 101, both of which are integers. In general, a magic rectangle can be created only if the number of rows and number of columns are both even or both odd; that is, a magic rectangle cannot be created if one is even and the other is odd. Of the four choices above, only 2 × 50 and 10 × 10 magic rectangles are possible. The following is a 10 × 10 magic rectangle (square) with a row and column sum of 505; other examples can be found online.

154

100

2

3

94

6

95

97

8

9

91

11

89

88

17

85

16

14

83

82

20

21

79

78

27

75

26

24

73

72

30

40

32

63

34

66

65

37

68

69

31

50

42

53

54

46

55

57

58

49

41

51

59

48

47

56

45

44

43

52

60

70

62

33

64

35

36

67

38

39

61

71

29

18

87

25

86

74

23

12

80

81

19

28

77

15

76

84

13

22

90

10

92

93

4

96

5

7

98

99

1

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Creating a 2 × 50 magic rectangle is left as a challenge for the reader.

PROBLEM 92. Fair and Square Common Core State Standards for Mathematics: 2.NBT.B.6 • 3.NBT.A.2 • MP.8

In the Classroom It’s educational malpractice for teachers to distribute worksheets that contain a large collection of indistinguishable exercises. Leinwand (2019) goes so far as to set a limit for the number of exercises a student should see. When students are doing more than four exercises on a mindless practice worksheet without pausing to get feedback on their work and to discuss alternatives and explanations, their learning is being compromised.

Practice is important so students develop automaticity and computational fluency, but there are superior ways to attain those skills. Problems like this one provide that opportunity. A 4 × 4 magic square contains 16 numbers, and the sum of every row and column is the same. To create a 4 × 4 magic square with a magic sum of 100, students might have to do a lot of exploration before seeing patterns. Throughout that process, they’ll perform potentially hundreds of addition calculations, providing computational practice that’s embedded in a problem with deeper mathematical purpose. To introduce this problem, present a 4 × 4 magic square to students that uses the numbers 1 to 16. Instead of presenting it as a magic square, ask them to speculate why mathematicians might refer to such a square as “magic.” Allow students some individual think time, and then have them share their ideas with a partner. That alone is usually enough for students to start considering sums, and immediately they are engaged in addition practice. From there, some students will proceed to finding a set of 16 numbers that would give a magic sum of 100, and then using trial-and-check to attempt to fill in the square; a few students will instead try to determine how they can modify the numbers in the standard magic square to obtain a magic sum of 100. In either case, students simultaneously engage in problem solving and computational practice.

Answer and Solution There are many 4 × 4 magic squares with a magic sum of 100. One possibility is to modify the famous magic square from Albrecht Durer’s painting Melancholia. It’s a standard 4 × 4 magic Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

155

square in which the sum of every row, column, and diagonal is 34. If each of the 16 numbers in that magic square were doubled, then the magic sum would be 68; that is, the magic sum would double. On the other hand, if each of the 16 numbers were increased by 2, the new magic sum would increase by 4 × 2 = 8 because there are four numbers in each row and column. With that background, is it possible to multiply each number by some factor and then add some amount to each number to change the magic sum from 34 to 100? Indeed, it is. A little trial-andcheck reveals that if each number is doubled and then increased by 8, the magic sum will increase to 2 × 34 + 4 × 8 = 100. 16

3

2

13

40

14

12

34

5

10

11

8

18

28

30

24

9

6

7

12

26

20

22

32

4

15

14

1

16

38

36

10

→

An equally valid, though significantly less satisfying, solution can be found using one of the many magic square creators that exist online. The following solution was obtained with one of them: 18

33

21

28

23

26

20

31

29

22

32

17

30

19

27

24

PROBLEM 93. Magic Sum Common Core State Standards for Mathematics: 3.NBT.A.2 • 6.SP.B.5.C • MP.1

In the Classroom The allure of magic squares has captured students for generations, but some of the allure is lost when students have to create magic squares on their own. Adkins (1963) notes that students “often lose interest before they successfully complete a magic square because of the frequency with which they must either erase their numerals or redraw their diagrams” (p. 498). He offers a suggestion, borrowed 156

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

from Henry Ernest Dudeney, that small squares of paper or cardstock can have the numerals written on them, and students can manipulate the squares into a potential arrangement, check the result, and then make modifications quickly. The benefit of allowing students to attempt to create a 10 × 10 magic square—which is not easy, by the way!—is to appreciate the perseverance required to be successful. In addition, multiple failed attempts will enlighten students as to the desired magic sum. For instance, if students create an arrangement for which the sums of the rows are 527, 513, 519, 544, 409, 527, 518, 495, 514, and 486, they may start to realize that some rows need to be increased and some need to be decreased, which eventually may lead to the recognition that the sum of each row must be the mean of those 10 sums.

Answer and Solution 505. There are 10 rows and 10 columns in a magic square with 100 numbers, and the sum of the numbers 1 + 2 + 3 + 4 + ⋯ + 100 = 5,050. Consequently, the magic sum will be 5,050 ÷ 10 = 505.

PROBLEM 94. One and Done Common Core State Standards for Mathematics: 3.OA.A.3 • 3.OA.C.7 • MP.1

In the Classroom To set the stage for this problem, organize a competition in your classroom. Students could play rock-paper-scissors, mancala, Nim, or some other mathematical game that can be completed quickly. Let students compete in a single-elimination tournament, then ask, “How many individual matches were needed to determine a champion?” It’s not uncommon for students to attempt to solve this problem with brute force rather than logical thinking, and students will likely try to count the number of matches that occurred. When students have found that number, encourage them to find the number of matches needed for a different number of students. Eventually, students will realize that n − 1 matches are needed for a single-elimination tournament with n students, and groans of “Oh, of course!” are likely when students experience that aha moment. As a follow-up, allow students to participate in a double-elimination tournament. Before students begin play, ask, “How many matches will be needed to determine a champion?” Have them compare and contrast this situation to the single-elimination tournament. An extension question is to ask how many rounds of play are required to determine a champion. If the tournament consists of n players, then k rounds will be needed, where k is the least positive integer such that 2k > n. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

157

Answer and Solution 99. One way to think about this problem is to consider the number of players who are eliminated each round. Round

Matches

Players Eliminated That Round

Players Remaining in Tournament

1

50

50

50

2

25

25

25

3

12

12

13

4

6

6

7

5

3

3

4

6

2

2

2

7

1

1

1

The total number of matches is 50 + 25 + 12 + 6 + 3 + 2 + 1 = 99. Another way to think about it is to realize that one player will be eliminated in each match, and 99 players must be eliminated to determine a champion. Therefore, 99 matches must occur.

PROBLEM 95. To the Point Common Core State Standards for Mathematics: 3.OA.A.3 • 8.EE.C.7 • MP.4

In the Classroom Students may attack this problem with an organized chart, trial-and-check, or using an equation. Students who have studied algebra formally, however, may abandon the equation under the assumption that one equation with two unknowns can’t be solved. In fact, in the vernacular of Karp, Bush, and Dougherty (2014), this is a “rule that expires.” It’s not that the equation can’t be solved; rather, there are many solutions. Algebraically, this situation can be represented by the equation 8x + 13y = 100. Students can use this equation to determine possible solutions. One option is to solve for x to get x = 100 −8 13y , and students can then consider what happens when various values are substituted for y. In one sense, this is a modified trial-and-check approach, but the rewritten equation makes it obvious that only even values for y could possibly yield a solution. For odd values of y, the expression 100 − 13y will also be 158

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

odd, and dividing by 8 will not yield an integer. Moreover, y < 8, since 13 × 8 = 104. Consequently, the insight from the equation reduced the possible values for y to 2, 4, and 6. When students begin to study algebra, it’s not uncommon for other solution strategies to be abandoned in favor of symbolic manipulation. That is a mathematical tragedy. Instead, solving equations should represent just one more tool in an already overflowing toolbox.

Answer and Solution Six times. The only combination of 8s and 13s equal to 100 is 6 × 8 + 4 × 13 = 48 + 52.

PROBLEM 96. Call to Order Common Core State Standards for Mathematics: 4.NBT.A.2 • 4.OA.C.5 • HSS.CP.B.9 • MP.7 • MP.8

In the Classroom The CCSSM standards listed above indicate quite a range, from fourth grade through high school, but any student comfortable with five-digit numbers could have success with this problem. Although they might need a definition and an example of permutations, students in upper elementary grades could otherwise handle the mathematics of this question. Older students who are already familiar with permutations will have a much easier time solving the problem, however, because they will be able to determine that there are 120 possible permutations, and there are 24 permutations that start with each digit 1 to 5. The following tasks could be used to extend this problem: n

 ow many five-digit integers could be formed with the digits 1, 2, 3, 4, and 5, if digits can be H repeated? If they are written in order from the least to the greatest, what number would be in the 100th position in the list?

n Have all students in the class create a permutation of the digits 1, 2, 3, and 4, and then have

students organize themselves in order of their permutations from the least to the greatest. Conduct a discussion about whether they’ve listed every possible permutation; if not, which ones are missing?

Answer and Solution 51,342. There are 24 permutations that start with each digit from 1 to 5. Consequently, the last number that starts with the digit 4, which is 45,321, will be in the 96th position. So then, it Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

159

just becomes a matter of determining the next four numbers in the list: 51,234, 51,243, 51,324, and 51,342.

PROBLEM 97. Triangles on a Grid Common Core State Standards for Mathematics: 8.F.A.3 • HSS.CP.B.9 • MP.6

In the Classroom This problem is ridiculously difficult. It’s okay to admit that to students, and it’s fair to recalibrate expectations. Students should be expected to make progress and identify patterns but identifying every possible case and performing all calculations without an error may be unreasonable. Given how many different cases have to be considered, this is an appropriate problem for a group problem-solving activity. In the solution below, an effective solution strategy is to consider the total number of sets of three points from the grid, and then subtract the number of sets of three points that are collinear. Consequently, students can work collectively to identify the types of collinear sets that exist. In the solution below, they are organized by slope, but it is absolutely appropriate for students to organize the collinear sets by some other criteria. The goal could be for the entire class to collectively find the answer to the problem.

Answer and Solution 97.2 percent. A 10 × 10 lattice contains 100 points, so there are 100C3 = 161,700 possible ways to choose three points at random. That’s the easy part. The hard part is determining how many of those combinations form a triangle. It’s actually easier to count the number of ways that three points can be chosen that won’t form a triangle, and then subtract. Three points will not form a triangle if they are collinear. Consider a line segment drawn through each row of points in the lattice. Those segments are horizontal, so their slopes are zero, and each segment passes through 10 points. These 10 segments are represented in the top row of the table below, and they account for 10 × 10C3 = 1,200 sets of collinear points.

160

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Similarly, a line segment can be drawn through each column of points in the lattice. There are 10 parallel segments that are vertical, so their slopes are undefined, and each segment passes through 10 points. These 10 segments are represented in the second row of the table, and they also account for 10 × 10C3 = 1,200 sets of collinear points.

Segments through n Points

Slope

3

4

5

6

7

8

9

10

0

—

—

—

—

—

—

—

10

undefined

—

—

—

—

—

—

—

10

1, −1

4

4

4

4

4

4

4

2

1 2

1 2

16

16

48

—

—

—

—

—

3, –3, , – 

1 3

1 3

72

28

—

—

—

—

—

—

1 1 4 4 2 2 3 3 ,− , ,− 3 3 2 2 3 3 4 4 , − , ,− 4 4 3 3

64

—

—

—

—

—

—

—

64

16

—

—

—

—

—

—

32

—

—

—

—

—

—

—

2, −2, , – 

4, – 4, , – 

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

161

Life gets a little more difficult when attempting to count the sets of three points that lie on a line segment that isn’t horizontal or vertical. The image below shows all segments with a slope of 1.

In that image there are two segments through three points, two segments through four points, two segments through five points, two segments through six points, two segments through seven points, two segments through eight points, two segments through nine points, but only one segment through 10 points. Moreover, there are the same number of segments with slope −1. A segment through three points accounts for 3C3 = 1 set of collinear points that won’t form a triangle, a segment through eight points accounts for 8C3 = 56 sets of collinear points that won’t form a triangle, and in general, a segment through n points accounts for nC3 sets of collinear points that won’t form a triangle. Therefore, the total number of sets of three points that won’t form a triangle and lie on a segment with slope 1 or −1 is as follows: 4 × (3C3 + 4C3 + 5C3 + 6C3 + 7C3 + 8C3 + 9C3) + 2 × 10C3 = 1,080 As represented in the table, there are many other segments that pass through anywhere from three through 10 points. On your own, you can verify that the total number of sets of three points that won’t form a triangle is 4,448. Since there are 161,700 ways to choose three points from a 10 × 10 lattice, and if 4,448 of those ways won’t yield a triangle, then 161,700 − 4,448 = 157,252 triangles are possible. The probability that a triangle will be formed by three randomly selected points from the 10 × 10 grid is 157,252 ≈ 97.2 percent. 161,700

PROBLEM 98. Binary Quandary Common Core State Standards for Mathematics: HSS.CP.B.9 • MP.1 • MP.7

162

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

In the Classroom This problem strays far from the path of traditional curriculum. Therein lies its beauty. The K−8 Publishers’ Criteria claims, “Problems are problems because students haven’t yet learned how to solve them” (CCSSO 2012, p. 16). This problem won’t assess a student’s ability to find the value of an expression, solve an equation, or graph a function. But it will require students to consider alternative problem-solving strategies. One strategy to use in the classroom is a combination of “solving a simpler problem” and “acting it out” strategies. Instead of a grid with 100 squares, place a 3 × 3 grid at the front of the room instead. To engage students, pass a coin around the room, and let students flip it. When the coin is flipped, place a 1 in a cell if the result is heads, 0 if the result is tails. The student who flips the coin gets to determine the cell in which the number is placed. After a student flips the coin, they should pass it to another student at an adjacent desk with the condition that they can’t give it to someone who has already flipped the coin. Continue in this manner until it’s not possible to use a random result. For instance, if the first two cells in the top row are filled, students should realize that the value in the last cell of the top row is uniquely determined: It’s a 1 if the sum of the previous two cells is odd; it’s a 0 if the sum of the previous two cells is even. The remainder of the 3 × 3 grid can be filled this way, with a brief discussion occurring after each flip about whether any additional cells are now uniquely determined. Cells that were randomly filled by the result of the coin toss should somehow be indicated, possibly by circling the number in that cell. The result of this process will show that once four cells have been filled by random flipping, the other five are uniquely determined. Repeat the process using a blank grid. It’s likely that different cells will be filled randomly, but once again, when four cells have their values randomly determined, the values in the other five cells are uniquely determined. From there, students should be able to extrapolate to determine the answer with a 10 × 10 grid.

Answer and Solution 281. This is a daunting problem, and one thought may be to consider all the cases. Two cases are relatively easy to consider: (1) If the sum of every row is 10, then every square is filled with a 1 and (2) if the sum of every row is 0, then every square is filled with a 0. A slightly more complex case is if every row has a sum of 6 and every column has a sum of 8. It would certainly be possible to count these possibilities using combinatorics, but it wouldn’t be easy. Moreover, what about the case where five rows have a sum of 6, four rows have a sum of 4, and one row has a sum of 2? The first question is whether or not that’s even possible—it is; see below for one example—but then, how would you count all of the possibilities? Surely, there’s got to be a better way. Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

163

1

1

1

1

1

1

0

0

0

0

1

1

1

1

1

1

0

0

0

0

1

1

1

1

1

1

0

0

0

0

1

1

1

1

1

1

0

0

0

0

1

1

1

1

1

1

0

0

0

0

1

1

1

1

0

0

0

0

0

0

1

1

1

1

0

0

0

0

0

0

1

1

1

1

0

0

0

0

0

0

0

0

0

0

1

1

1

1

0

0

0

0

0

0

0

0

1

1

0

0

An alternative approach is to consider a simpler case. Let’s look at a 2 × 2 grid and see what happens. Clearly, there are only two possibilities: All cells can be filled with 1s, or all cells can be filled with 0s. 1

1

0

0

1

1

0

0

Why are there only two possibilities? Well, if the upper left cell is filled with a 1, then the cells to its right and below it must also be filled with 1s to get an even sum in the top row and first column; and then, the lower right cell must be filled with a 1 to get an even sum in the bottom row and second column. A similar argument follows if the upper left cell is filled with a 0. What about a 3 × 3 grid? As it turns out, there are 16 possibilities, which I won’t draw here. But let’s see what happens when we start to fill in the grid. Let’s say we start with a 1 in the upper left cell; does that then dictate what happens in other cells? No, we are still free to choose randomly. So, let’s put a 0 in the middle cell of the top row; does that dictate what happens in other cells? Yes! The upper right cell must now be a 1, so that the top row has a sum of 1 + 0 + 1 = 2. Then, let’s put a 1 in the first cell of the middle row; does that dictate what happens in other cells? Yes, again! The lower left cell must be a 0, so that the first column has a sum of 0 + 1 + 1 = 2. Now let’s put a 0 in the middle cell; does that dictate what happens in other cells? Absolutely! In fact, it dictates what must happen in each of the three remaining cells: a 1 must be placed in the middle cell of the last column, to give 1 + 0 + 1 = 2 in the middle row; a 0 must be placed in the middle cell of the bottom row, to give 0 + 0 + 0 = 0 in the second column; and, finally, a 0 must be placed in the lower right cell to give 1 + 1 + 0 = 2 in the third column.

164

1

0

1

1

0

1

0

0

0

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

In the figure above, the 0s and 1s that were placed randomly are in black, but the 0s and 1s that were dictated by other choices are in gray. Since we randomly placed 0s and 1s in four cells, then there must be 24 = 16 ways to fill the grid with an even sum in every row and column. More generally, once the first n − 1 rows and n − 1 columns of an n × n grid are randomly filled with 0s and 1s, then the entries in the remaining 2n − 1 cells will be dictated. For a 10 × 10 grid, this means that nine rows and nine columns can be filled randomly, and there are 281 ways to do that.

PROBLEM 99. Squares on a Grid Common Core State Standards for Mathematics: 5.G.B.3 • 6.G.A.3 • MP.7

In the Classroom A few students will realize that the number of squares of a given size can be counted quickly. For instance, some students will recognize that there are eight rows of eight squares of size 3 × 3, so there must be 8 × 8 = 64 three-by-three squares. Most students, however, will attempt to count the squares one-by-one, which will require more time as well as lead to some errors. Therefore, assign squares of various sizes to different groups of students. Or better yet, allow students to choose which squares they’d like to count. Because the 1 × 1 and 10 × 10 squares are easiest to count, already have those values listed in a two-column table as follows: Size of Squares

Number of Squares

1×1

100

2×2 3×3 4×4 5×5 6×6 7×7 8×8 9×9 10 × 10

1

Students should verify their results with a classmate—and then with you—before a number is recorded in the table. When the right column is complete, a discussion about the resulting pattern,

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

165

as well as strategies for finding the sum, should occur. With some rearrangement and regrouping, finding the sum using mental math is possible as shown below: (1 + 4) + (9 + 16 + 25) + (49 + 81) + (36 + 64) + 100 = 5 + 50 + 130 + 100 + 100 = 385

Answer and Solution 385 squares. Obviously, there are one hundred 1 × 1 squares. Several 2 × 2 squares have been shaded below, but how many 2 × 2 squares are there in total? Well, every interior intersection of two line segments could be the center of a 2 × 2 square, and there are 81 such intersections. Consequently, there are eighty-one 2 × 2 squares.

In general, there are (11 − n)2 squares in the figure, for n = 1, 2, 3, . . ., 10. In total, that gives 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385 squares. (See problem 24, Square Deal, for a formula to calculate that sum.)

PROBLEM 100. Covering with Squares Common Core State Standards: 8.EE.A.2 • HSG.MG.A.3 • MP.7

In the Classroom It’s not uncommon for difficult problems to require the collective effort of many. For instance, when trying to find three cube numbers that add to 42 (see problem 56: Sum of Cubes), a combined team of professors from the University of Bristol and the Massachusetts Institute of Technology harnessed the computing power of 500,000 idle, unused home computers to generate a solution. Even with that kind of power, it still required over 1,000,000 hours to find the elusive answer. 166

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

In a similar but slightly less daunting way, this problem lends itself to collective problem solving. In the classroom, write the numbers 1 to 100 on a whiteboard or piece of chart paper, or simply display a laminated hundreds board. This will be used as a class recording sheet when arrangements for various numbers are found. Then, present the problem to students, showing the examples for 100, 25, and 13 squares. (Those numbers were chosen intentionally for the problem stem. The first two emphasize that repeated use of the same square is required; the last shows that arrangements may lack symmetry or regularity.) Ask every student to individually find one arrangement of squares that will cover the 10 × 10 grid. Do not let students shout their answers; instead, ask students to put their fist in front of their chest when they have an arrangement they’re willing to share. Circulate as students work to ensure that every student has found at least one arrangement; in addition, encourage students who have already found an arrangement on their own to find others. As you circulate, make a mental note of the order in which you’d like students to share their arrangements. This is a very good opportunity to engage reluctant learners. If you notice that some typically tentative students have found an arrangement, ask them to share first to encourage their participation in the class discussion. As arrangements are shared and validated, cross those numbers off the recording sheet. The collective power of the class will likely reduce the list to 85 or fewer. As a note of caution, some students will notice that many numbers can be removed simultaneously. For instance, the arrangement of 25 squares shown in the problem stem can be successively modified by converting each 2 × 2 square into four 1 × 1 squares, and this yields arrangements for 25, 28, 31, . . ., 3n + 1, . . ., 100 squares. It’s important to prevent students from sharing this insight too early as it will diminish the problemsolving opportunity for the rest of the class. After every student has shared at least one arrangement, divide the class into groups to find other arrangements. During these small-group discussions, patterns of arrangements will be discovered. When all groups have made progress, reconvene the class for a large-group discussion. At this point, it is appropriate to have students share their insights about how to modify arrangements to eliminate many numbers from the recording sheet. Following this discussion, it’s not uncommon for 70 to 80 numbers to have been eliminated from the recording sheet. Finding an arrangement for the remaining 20 to 30 numbers can be difficult, however. Additional class time can be allocated for this, but an alternative is to record the numbers for which an arrangement has not been found in a prominent place in the classroom. Tell students that they can continue to work on the problem outside of class, and when a student shows up on a subsequent day with another arrangement found, make a big deal out of it! Share the arrangement with all students, and let the student who made the discovery cross the number off the recording sheet. About a week later, it would be appropriate to use a few minutes of class time to finalize the discussion.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

167

Answer and Solution It’s not possible to cover a 10 × 10 grid 2, 3, 5, 6, 7, 9, 87, 90, 93, 95, 96, 98, or 99 squares. It’s obviously possible to cover the entire grid with one hundred 1 × 1 squares. But if four 1 × 1 squares are converted to a 2 × 2 square, as shown below, the number of squares is reduced to 97. Convert four more 1 × 1 squares to a 2 × 2 square, and the number of squares is reduced to 94. Continue in this manner, and the entire grid can be covered with 91, 88, 85, . . ., 25 squares.

In a similar manner, if the board is covered with one hundred 1 × 1 squares, and nine squares are successively converted to a 3 × 3 square, the number of squares is successively reduced by eight squares each time. Hence, the entire grid can be covered with 92, 84, 76, . . ., 28 squares, though some of those were already counted above. The board can be covered with a 3 × 3 square and ninety-one 1 × 1 squares, but if four of the 1 × 1 squares are converted to a 2 × 2 square, the number of squares is reduced by three. Hence, the entire grid can be covered with 92, 89, 86, . . ., 29 squares. Cover the board with two 3 × 3 squares and eighty-two 1 × 1 squares, for a total of 84 squares. If four of the 1 × 1 squares are then converted to a 2 × 2 square, the entire grid can be successively covered with 84, 81, 78, . . ., 24 squares. The board can be covered with a 6 × 6 square and sixty-four 1 × 1 squares, for a total of 65 squares. But if four 1 × 1 squares are converted to a 2 × 2 square, the number of squares is reduced by three. Hence, the entire grid can be covered with 62, 59, 56, . . ., 17 squares. The board can be covered with one 3 × 3 square, one 2 × 2 square, and eighty-seven 1 × 1 squares, for a total of 89 squares. Once again, successively convert four 1 × 1 squares to a 2 × 2 square to cover the grid with 86, 83, 80, . . ., 29. The board can be covered with one 8 × 8 square and nine 2 × 2 squares, for a total of 10 squares. Successively dividing each 2 × 2 square into four squares will cover 13, 16, 19, . . ., 37 squares.

168

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

At this point, many numbers have been eliminated, and they are shown in gray below: 1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

The arrangement below consists of one 6 × 6 square, three 4 × 4 squares, and four 2 × 2 squares, for a total of eight squares. The 2 × 2 squares can be successively divided to cover the entire grid with 8, 11, 14, or 17 squares.

The following figure uses 1 × 1, 2 × 2, 3 × 3, and 5 × 5 squares to cover the grid with 12 squares. If the 2 × 2 squares are successively divided, the grid can also be covered with 15, 18, or 21 squares.

Solutions and Suggestions

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

169

Finally—and obviously—the square can be covered with four 5 × 5 squares.

PROBLEM 101. Falsehoods and Fibs Common Core State Standards for Mathematics: MP.3

In the Classroom While not terribly mathematical, this problem encourages problem solving and logical thinking. It works well as a warm-up that can be used while taking roll call, or it could be posed as a problem of the week. Parker (1955) noted that puzzles can be used in the classroom to serve two purposes: to secure the interest and attention of the group and to illustrate and clarify certain mathematical concepts and techniques. The technique illustrated by this problem is deductive reasoning. Reasoning is inherent in developing the components of mathematical competence (National Research Council 2001). The important purpose of the puzzle is to require students to justify their answer through deductive reasoning.

Answer and Solution Statement 99: Consider statement 1. If exactly one statement in this list is false, then 99 must be true. So, statement 1 cannot be true. Consider statement 2. If exactly two statements are false, then 98 are true. So, statement 2 cannot be true. And so on. Consequently, only one statement can be true; a contradiction would arise if more than one was true. If only one is true, then 99 must be false, so statement 99 is true.

170

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

References Adkins, Bryce E. 1963. “Adapting Magic Squares to Classroom Use.” The Arithmetic Teacher 10 (December): 498−500. Baker, Austin T., and Josh Cuevas. 2018. “The Importance of Automaticity Development in Mathematics.” Georgia Educational Researcher 14(2): Article 2. Berry, Wendell. 1983. Standing by Words: Essays. San Francisco: North Point Press. Bieda, Kristen N., and Megan Staples. 2020. “Justification as an Equity Practice.” Mathematics Teacher Learning & Teaching, 113 (February): 102−8. Brown, Peter C., Henry L. Roediger III, and Mark A. McDaniel. 2014. Make It Stick: The Science of Successful Learning. Cambridge, MA: Belknap Press of Harvard University Press. Butts, Thomas. 1980. “Posing Problems Properly.” In Defining Mathematics Education: Presidential Yearbook Selections 1926–2012, 75th Yearbook of the National Council of Teachers of Mathematics, edited by Francis (Skip) Fennell and William R. Speer. Reston, VA: National Council of Teachers of Mathematics. Champagne, Zak. 2020. “Playing the Long Game: Examining the Routines, Tasks, and Structures That Support Student Learning Over Time (Video webinar, February 10, 2020)].” In Effective Practices for Advancing the Teaching and Learning of K−5 Mathematics. Retrieved from http://mathleadership.org/effective-practices-2/. Council of Chief State School Officers CCSSO. 2012. K−8 Publishers’ Criteria for the Common Core State Standards for Mathematics. https://achievethecore.org/page/267/publisherscriteria-for-the-ccss-in-mathematics. Cooper Foreman, Linda, and Albert B. Bennett. 1995. Visual Mathematics: Course I. Salem, OR: The Math Learning Center. Retrieved from https://www.mathlearningcenter.org/sites/ default/files/pdfs/VM1-Guide-Web.pdf. Drake, Corey. 2017. Number Choice Matters (Blog post; May 8, 2017). Retrieved from https://www. nctm.org/Publications/Teaching-Children-Mathematics/Blog/Number-Choice-Matters/. Fey, James T. 1996. Bits and Pieces II: Using Rational Numbers. New York: Dale Seymour Publications. Finkel, Benjamin Franklin, and Colaw, James M. 1894. "Introduction." American Mathematical Monthly 1 (January): 1.

References

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

171

Fletcher, Graham. 2020. “Building Math Residue with Lessons That Stick (Video webinar; February 24, 2020).” In Effective Practices for Advancing the Teaching and Learning of K−5 Mathematics. Retrieved from http://mathleadership.org/effective-practices-2/. Hamilton, Gordon. 2018. “Picasso’s Line Puzzle.” Retrieved 17 February 2020 from https:// mathpickle.com/project/picassos-line-puzzles. Hawking, Stephen W. 2005. God Created the Integers: The Mathematical Breakthroughs That Changed History. Philadelphia: Running Press. Hawthorne, Casey, and Bridget K. Druken. 2019. “Looking for and Using Structural Reasoning.” Mathematics Teacher 112 (January): 294−301. Heath, Chip, and Dan Heath. 2010. Switch: How to Change Things When Change Is Hard. New York: Broadway Books. Hiebert, James, and Douglas A. Grouws. 2007. “The Effects of Classroom Mathematics Teaching on Students’ Learning.” In Second Handbook of Research on Mathematics Teaching and Learning, edited by Frank K. Lester Jr., pp. 371−404. Charlotte, NC: Information Age Publishing. Heitschmidt, Corey. 2008. “Fractional Clothesline.” Retrieved from https://www.nctm.org/ Classroom-Resources/Illuminations/Lessons/Fractional-Clothesline. Jaccard, Clarence R. 1956. “Objectives and Philosophy of Public Affairs Education.” In Increasing Understanding of Public Problems and Policies: 1956. Chicago: Farm Foundation. Jansen, Amanda, Brandy Cooper, Stefanie Vascellaro, and Philip Wandless. 2016. “RoughDraft Talk in Mathematics Classrooms.” Mathematics Teaching in the Middle School 22 (December): 304−7. Kalinec-Craig, Crystal A. 2017. “The Rights of the Learner: A Framework for Promoting Equity through Formative Assessment in Mathematics Education.” Democracy & Education 25 (2): 4−5. https://democracyeducationjournal.org/cgi/viewcontent. cgi?article=1298&context=home. Karp, Karen S., Sarah B. Bush, and Barbara Dougherty. 2014. “13 Rules That Expire,” Teaching Children Mathematics 21 (August): 18−25. Kasner, Edward, and James Newman. 1940. Mathematics and the Imagination. New York: Simon & Shuster. Leinwand, Steve. April 2019. “Great Math Leaders as Rebels with a Cause for Students.” Session at the National Council of Teachers of Mathematics Annual Meeting and Exposition, San Diego. Slides available online at http://steveleinwand.com/wp-content/uploads/2019/04/ NCSM-San-Diego-2019-Rebels.pdf. 172

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Leinwand, Steve. December 2015. “Breathing Classroom Life into the Teaching Practices.” Paper presented at the 2015 CMC North Conference, Asilomar, CA. Lockhart, Paul. 2002. A Mathematician’s Lament. Unpublished manuscript. Retrieved from https://www.maa.org/external_archive/devlin/LockhartsLament.pdf. Lollos, E. 2014. “Re: The Locker Problems Solved (Blog comment; November 26, 2014).” Natural Blogarithms. Retrieved from https://natblogarithms.wordpress.com/2011/04/28/the-lockerproblems-solved/#comment-905. Math Jokes 4 Mathy Folks. 2011. “An Open and Shut Case (Blog post; November 26, 2011).” Retrieved from https://mathjokes4mathyfolks.wordpress.com/2011/04/18/an-open-andshut-case/. Maubossin, Andrew, and Michael J. Maubossin. 2018. “If You Say Something Is, ‘Likely,’ How Likely Do People Think It Is?” (Harvard Business Review: July 3, 2018.) Retrieved from https://hbr.org/2018/07/if-you-say-something-is-likely-how-likely-do-people-think-it-is. McNally, Joe, and William Murray. 1964. The Ladybird Key Word List. Loughborough, UK: Wills & Hepworth. National Council of Teachers of Mathematics (NCTM). 2014. Principles to Actions: Ensuring Mathematical Success for All. Reston, VA: NCTM. _____. n.d. Beginning to Problem Solve with I Notice, I Wonder. Retrieved from https://www.nctm. org/Classroom-Resources/Problems-of-the-Week/I-Notice-I-Wonder/. National Governors Association Center for Best Practices (NGA Center) and Council of Chief State School Officers (CCSSO). 2010. Common Core State Standards for Mathematics. Washington, DC: NGA Center and CCSSO. http://www.corestandards.org. National Research Council. 2001. Adding It Up: Helping Children Learn Mathematics, edited by Jeremy Kilpatrick, Jane Swafford, and Bradford Findell. Washington, DC: National Academies Press. Naylor, Rhonda. n.d. “Supreme Court Handshake.” Retrieved from https://www.nctm.org/ Classroom-Resources/Illuminations/Lessons/Supreme-Court-Handshake/. Niederman, Derrick. 2009. Illuminations. “Brain Teasers: Sliding Triangle.” Retrieved March 3, 2020, from https://illuminations.nctm.org/BrainTeasers.aspx?id=4751. NRICH. 2011. “Creating a Low Threshold High Ceiling Classroom.” Retrieved from https:// nrich.maths.org/7701. Parker, Jean. 1955. “The Use of Puzzles in Teaching Mathematics.” The Mathematics Teacher 48 (April): 218–27. References

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

173

Paulos, John A. 1998. Innumeracy: Mathematical Illiteracy and Its Consequences. New York: Hill and Wang. Pólya, George 1945. How to Solve It. Princeton, NJ: Princeton University Press.  Reinhart, Steven C. 2000. “Never Say Anything a Kid Can Say.” Mathematics Teaching in the Middle School 5 (April): 478–83. Reiter, Harold, Arthur Holshouser, and Patrick Vennebush. 2012. “Don’t Fence Me In!” Mathematics Teacher 105 (April): 594–9. Seeley, Cathy L. 2014. Smarter Than We Think: More Messages About Math, Teaching and Learning in the 21st Century—A Resource for Teachers, Leaders, Policy Makers and Families. Sausalito, CA: Math Solutions. Shah, R. [Raj Shah]. 2020. Tips for Math at Home (Video File; March 25, 2020). Retrieved from https://www.youtube.com/watch?v=cHwdmu3y_nI. Shore, Chris. 2018. Clothesline Math: The Master Number Sense Maker. Huntington Beach, CA: Shell Educational Publishing. Singh, Simon. 2013. “The Whole Story.” Retrieved from https://simonsingh.net/books/fermatslast-theorem/the-whole-story/. Smith, Nanci N. 2017. A Mind for Mathematics: Meaningful Teaching and Learning in Elementary Classrooms. Bloomington, IN: Solution Tree. Smith, Margaret S., and Mary Kay Stein. 2011. 5 Practices for Orchestrating Productive Mathematics Discussions. Reston, VA: National Council of Teachers of Mathematics. Sofroniou, Anastasia, and Konstantinos Poutos. 2016. “Investigating the Effectiveness of Group Work in Mathematics.” Education Sciences 6 (3) 30: 1–15. Retrieved from doi:10.3390/ educsci6030030. Stadel, Andrew. 2014. Estimation 180 Handout. http://www.estimation180.com/blog/estimation180-handout. Stuart, Morag, Maureen Dixon, Jackie Masterson, and Bob Gray. 2003. “Children’s Early Reading Vocabulary: Description and Word Frequency Lists.” British Journal of Educational Psychology 73 (January): 585–98. Tanton, James. 2013. Without Words: Mathematical Puzzles to Confound and Delight. London: Tarquin Press. Tobias, Sheila. 1993. Overcoming Math Anxiety. New York: W. W. Norton & Company.

174

ONE HUNDRED PROBLEMS INVOLVING THE NUMBER 100

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

Trémolière, Bastien, and Wim De Neys. 2014. “When Intuitions Are Helpful: Prior Beliefs Can Support Reasoning in the Bat-and-Ball Problem.” Journal of Cognitive Psychology 26 (March): 486–90. https://www.psychologicalscience.org/publications/observer/obsonline/a-newtwist-on-a-classic-puzzle.html. von Mises, Richard. 1964. Selected Papers of Richard von Mises: Probability and Statistics, vol. 2, edited by Philipp Frank, Sydney Goldstein, Mark Kac, William Prager, Gabor Szegö, and Garrett Birkhoff. Providence, RI: American Mathematical Society, 313–34. Wolfram, C. 2010. "Teaching Kids Real Math with Computers" [Video file]. Retrieved from https://www.ted.com/talks/conrad_wolfram_teaching_kids_real_math_with_computers. Zucker, Joshua. 2017. “Problem Solving.” Paper prepared for Wayne and Washtenaw Counties (Michigan) Math Teachers’ Circle. Retrieved from http://www.math.lsa.umich.edu/ WCMTC/ProblemSolving_Zucker.pdf. Zucker, Joshua. 2012. “Be Less Helpful.” MTCircular (August 2012): 4–7. Retrieved from https:// www.mathteacherscircle.org/assets/legacy/newsletter/MTCircularAutumn2012.pdf.

References

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.

175

One Hundred Problems Involving the Number 100 Share the wonder, joy, beauty—and fun— of mathematics with these 100 problems. The problems in One Hundred Problems Involving the Number 100 celebrate the “Goldilocks” charm of 100—a number not too small, not too large, but just right to challenge students without intimidating them. It’s used in myriad ways within the problems: as an exponent, product, area, or perimeter; a constant in an equation; the number of items in a series or sequence; or as a physical value, such as a stack of 100 coins, a deck with 100 cards, or a jug that holds 100 ounces. There is something magical about the number 100, and these problems aim to capture some of that magic. But no matter how 100 is used, each problem is meant to spark curiosity and motivate students (and their teachers) to want to solve it. One Hundred Problems Involving the Number 100 is not just a list of problems. n The

problems are designed to promote classroom discourse, allow students to think deeply about mathematical concepts, and learn problem-solving strategies, as well as to make connections between different topics in mathematics.

n The

purpose of the problems is to promote a variety of problem-solving strategies. There is no one “right way” to solve these tasks.

n A

range of mathematical topics from patterns, conversions, and sums and series to number theory, functions, probability and statistics, and geometry are covered.

n

The problems are aligned to identified Common Core State Standards for content and math practices. The chart in Part 3 lets teachers find problems at an appropriate level of difficulty for their students.

n

For each problem, you’ll find a description of how the problem might be used in the classroom, suggestions for how to provide assistance to students without divulging the answer or even exposing a solution strategy, and possible extensions.

You’ll find both practical information and inspiration in these pages as well as a treasure trove of meaningful mathematical tasks to engage and excite your students.

Copyright © 2020 by the National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in other formats without written permission from NCTM.