Exerciţii şi probleme de algebră, combinatorică şi teoria numerelor
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- Author / Uploaded
- Dragoş Popescu
- George Oboroceanu
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;c>in ·ecuafiile (2) �i ,(i) �i ;poi din ecuatiil� ( 1) �i (3) rezulti!.:
E
I0kci, un numar natural scris în sistemul z�èima'i, �i s(n) suma cifrelbr k=O sale. Avem: n = s(n) (mod 9). Într-adevar: 11.5. Fie n =
n=E p
k=O
107,,k =
p p . . E (9+ I)kc = E 'k = s(n) (mod 9). k=O k
k=O
Deoarece' s(n) �n (egalitate numai în ca�ul în care n are o stngura c,ifra) rezulta ca �rul n, s (n), s(s(n)), .,. este descrescator �i devine stationar de la ·un .am:imit rang încolo, egal fiind eu restul lui. n modulo 9. Réve�nd la problema, observam ca �irul de numere de o cifra construit este compus din repetarea secvenfei 1, 2, ... , 9; cürele 1 �i 2 apar mereu alaturi. Deoarece 1 000 000= 1 (mod !t, rezulta ca numarul 'cifrelor 1 depa�ewte eu o unitate numarul cifrelor 2. ll.6. Fie n = xyz un numar.de tr.ei cifre. Daca el contine cifre nule atunci pri� suma pa tratelor cifrelor sale se obtine, de fapt, suma patratelor cifrelor unui numar de doua cifre. Daca l�x. y, z�9 atunci: x2
+ y2 + z2 < 100 x +
10:,, + z.
Într-adevar, tinînd cont ca z(z -1) � 72 �i y{y -10) < 0 pentru I�y. z�9. avem: x2. - 100 x +..y(y - 10)
+ �(z - 1) < x2 -
lOOx
+ 72 < O.
Rezulta a�ar, ca este suficient sa demonstram proptj.etatea din enunt pentru numereie eu doua cüre, fapt care rezulta relativ u�or prin analizarea directa a numerelor deoarece multe cazuri se reduce unele la altele. 11.7. Solu/ia '1. În primele 200 çle num:re naturale 100 sînt pare, 100 impare. Fie a1,a2, ••• .., ai.01 null)erele alese. • Avem 'li = 2111 h1, a2 = 2112 h2, ••.•, a101 = 2cx,., h101 , undè ht sînt impare �i 1Xt;;;,, 0, V i, l�i�·IOL cum' h1 < 200 �i cum au fost alese 101 numere · impare rezulta ca exista o pereche (i, j) as�el încît h, = h1 ·, Daca IX! > a.1 rezulta a1 1 a, iar daca.1Xi > IX! rezuJta a, l a1, q.e.d. Evident, problema :,e poate generaliza pentru �irul' de numere naturale 1, 2, 3, ..., 2n din care se aleg n + I numer·e, _soh\j:ia proble!11ei fiin� a.naloaga. ,1 Solufia 2. În çazul general problema pe;vute o 'solutie prin_inductie: « Fie 1, 2, 3, ..., 2n nuniere naturale. Daca alegem n + I ·dintre ele, atunci în acestea exista o pereche în care. un numar este divizorul c1;luilalt ,,. Pentru n = 1, 2, 3, banal. Sa presuppnem enuntul adevarat pentru n �i sa consideram �irul: 'l, 2, 3, ..., 2n, 2n
+ 1, 2n + 2.
1) Daca ;;irul a1 , ••. , an, an+i este ales din 1, 2; ..., 2n. se .aplica ipoteza ipducfiei: . 2) În'caz ,contrar,' avem an+i = 2n + 1 ;;i an+2 = 2(n + 1). Daca n + le{"a1 , a2, ••• , a.n+2 } atunci n + l l an+2 ·.
.
15 - Exerciîii �i probleme de �lgebra - c. 2580 ·
225
'
+
Daeà n le{a1, a2, •••, a,.+2} adaugàm aeest numar �i formam un sub�ir den+ l numere. al �rului. Conform ipotezei induetiei 3a11:, a1( l :s;; k :s;; l :s;; 2n astfel îneît a1c I ai sau 3a1c(1 :s;; :s;; k :s;;2n) astfel încît a1,; 1 n+ · l; in aeest . din urma. eaz avem a1c I a,.+t deoareee n+ l I a..+2 = = 2(n l).
+
11.8. Presupunem eà àlegem 100 de numere mai mici decît 200 astfel ea nici unul sa nu-1 dividà pe altul. Aràtam ea in aeest eaz nici unul nu poate fi mai mie decît 16. Consideràm cei mai mari divizori impari ai numerelor alese. Aee�tia sînt nunierele impare [x ]eM=>[[AY]eM=> ...
Dar: 1 2
!, ...
[ ¾ ] 2 [ x ] :s;;x 2k --+ Deoarece x li>➔ C:O
rezulta ea exista k0 astfel încît:
¾
1
1
[[ . . [[JJ t J2]
= 1 pentru k;., k0•
k ori
b) Observam acum ca. puterile lui 2 apartïn lui M. Într-adevir: leM=>2 2 eM=:,.2 4eM => .... Deci 2zteM(k = 0, 1,2, ...). Dar atunei [.J2 2"'] = 2"eM (k = o, 1,2, ...) e) .Sa consideram un numar oareeare neZ+ · Pentr.u a demonstra eà ne M este sufi
cient sa arâtà.m eà 3ke N astfel încît n2 eM. "'
Într-adevàr, daeâ exista un astfel de k atunei:
[V--] k 1 2k = n2 - eJl.f n
� succesiv obtinem ne M. d) Pentru ociœ keN exista P1ceN astfel foeît: ., 2Pt ._;n 2" < 2Pt+1 (unde:
226
p" = [lg2 n2
k
]).