Răspuns:
n se divide cu 7
Explicație pas cu pas:
n = 1+2+2²+2³+2⁴ + ... + 2²⁰¹¹ + 2²⁰¹² se divide cu 7
Aceasta adunare daca o transformam intr-un produs de factori si un factor este 7, atunci numarul n se divide cu 7.
Trebuie sa gasim un divizor 7 al lui n
n = 1 + 2 ( 1+2+2² +2³ +...+ 2²⁰¹⁰ + 2²⁰¹¹ )
n = 1 + 2 [ 1+ 2 ( 1+2+2² +...+ 2²⁰⁰⁹ + 2²⁰¹⁰ ) ]
n = 1+2 { 1 + 2 [1+2 (1+2 +...+ 2²⁰⁰⁸ + 2²⁰⁰⁹ )]}
Ce observam ?
Daca tot vom da factor comun pe 2, ultimii termeni ai adunarii se micsoreaza cu 1
Putem scrie:
n = 1+2 { 1 + 2 [1+2 (1+2 ····································)1 + 2· ( 1+2² + 2¹ )]}
n = 1+2 { 1 + 2 [1+2 (1+2 ·······························)1 + 2· [ 1 +2· (2+1)]}
n = 1+2 { 1 + 2 [1+2 (1+2 ·······························)1 + 2· [ 1 +2· (3)]}
n = 1+2 { 1 + 2 [1+2 (1+2 ·······························)1 + 2· [ 1 +6]}
[ 1 +6] = 7
Deci n se divide cu 7
