1. n: 7 = x rest 4n: 9 = y rest 9n: 63 = z rest aCat este a?2.abc numar natural de 3 cifre care impa.... Idea question from @Sabina

December 2018 1 381 Report

1. n: 7 = x rest 4
n: 9 = y rest 9
n: 63 = z rest a
Cat este a?

2.abc numar natural de 3 cifre care impartit pe rand la 3 numere naturale consecutive, obtinem caturi tot numere naturale consecutive, iar suma celor 3 resturi 23.

3. Se dau numerele A= $1^{2}$ + $2^{2}$ + $3^{2}$ + ...+ $2002^{2}$ si B= 1×3+ 2×4 + 3×5+ ... + 2001× 2003. Calculati A-B

VA ROOG AM NEVOIE DE ELE IN 2 ORE :(((

bunicaluiandrei

1. n=7x+4 n=9y+9 = 9(y+1) ⇒ n divizibil cu 9, adica, n = 9k ⇒ 9k-4 =7x ≡ (7k-7+2k+3)/7 =[(k-1) +(2k+3)/7]∈N ⇒2k+3 = 7q k= (7q-3)/2 , q=nr.impar (2k+3) ∈{7q} = {7,21,35,,,,,,,,,,(2α+1)·7} ⇒ k∈{ 2, 9, 16 ......(7α+2)} ⇒⇒ ⇒ n∈{18,81,144,.........9(7α+2)} ⇒18=63×0 +18, 81=63×1 +18 153 =63×2+18 ⇒ a=18

2. n= (x-1)·(q+2) + r1 (1) n=x(q+1) + r2 (2) n= (x+1)q+r3 (3) din (1)+(2)+(3) ⇒ 3n=xq-q+2x-2 +xq+x+xq+q +23 = 3xq+3x+21 din( 2)·3 ⇒3n =3xq - 3x +3r2 ⇒ 3xq-3x+21 = 3xq-3x+3r2 ⇒ r2=7 ⇒r1+r3 =16 n-r1 =(x-1)(q+2) n-r3 = (x+1)q r1-r3 = 2q -2x +2 =2(q-x+1) ⇒r1-r3 =nr. par ⇒ptr. r1-r3 =2 r1+r3=16 ⇒r1= 9 ; r3=7=r2 ptr r3≠r2 alegem r1-r3=4 ⇒r1=10 r3 = 6 ⇒ 4 =2 (q-x+1) q-x+1=2 ⇒q=x+1 ⇒n-10 =(x-1)(x+3) si n-6 = (x+1)(x+1) iar, deoarece n ≥ 100 ⇒ (x+1)² = 100 x=9 q = 10 n = 106 106 = 8×12 +10 106 = 10×10+6