Question

Calculati cate nr. naturale de cel mult trei cifre impartite la 30 dau restul 7 si apoi calculati suma acesto nr.

Answer 1

30xC+7=cel mai mare numar de 3 cifre, rezulta 30x33+7=997
Asadar C ia valori de la 1 pana la 33, deci avem 33 de numere.

Calculam suma lor:

C=1, 30x1+7=37
C=2, 30x2+7=67
C=3, 30x3+7=97
...
C=33, 30x33+7=997

37+67+97+...+997=(30+7)+(60+7)+(90+7)+...+(990+7)=
(30+60+90+...+990)+(7+7+...de 33 de ori...+7)=

30(1+2+3+...+33)+7x33=30x(33x34)/2 +7x33=

( am aplicat 1+2+...+n=n(n+1)/2  )

=15x34x33+7x33=(15x34+7)x33=517x33=17061

Answer 2

37 : 30 = 1  rest 7
67 : 30 = 2 rest 7
97 : 30 = 3 rest 7
127 : 30 = 4 rest 7
157 : 30 = 5 rest 7
187 : 30 = 6 rest 7
217 : 30 = 7 rest 7
247 : 30 = 8 rest 7
277 : 30 = 9 rest 7
307 : 30 = 10 rest 7
........................
997 : 30 = 33 rest 7
37 + 67 + 97 + .......... + 937 + 967 + 997 =
30 * 1 + 7 + 30 * 2 + 7 + ...... + 30 * 33 + 7 =
30 * (1 + 2 + 3 + ... + 33) + 7 * 33 =
30 * 33 * (33 + 1) : 2 + 7 * 33 =
30 * 33 * 17 + 7 * 33 =
16830 + 231 =
17061