Question

Demonstrati ca numarul 13^n+7^n-2 se divide cu 6, oricare ar fi n un numar natural.

Answer 1

$\boxed{(a+b)^{n}=M_{a}+b^{n}}$

$13^{n}+7^{n}-2=\\ \\ =(6+7)^{n}+7^{n}-2=\\ \\ =M_{6}+7^{n}+7^{n}-2=\\ \\ =M_{6}+2\cdot7^{n}-2=\\ \\ =M_{6}+2(7^{n}-1)=\\ \\ =M_{6}+2[(6+1)^{n}-1]=\\ \\ =M_{6}+2( M_{6}+1^{n}-1)=\\ \\ =M_{6}+2\cdotM_{6}=\\ \\ =M_{6}(1+2)=\\ \\ =M_{6}\cdot3,~divizibil~cu~6$