Question

Aratati ca numarul A=7+7la puterea a 2+7la putereaa 3+...+7la puterea 100 este divizibil cu 50

Answer 1

Combini termenii 2 cate 2: cei impari si cei pari, in mod consecutv
$A=(7+7^{3})+(7^{2}+7^{4})+(7^{5}+7^{7})+(7^{6}+7^{8})+..+(7^{97}+7^{99})+(7^{98}+7^{100})=7(1+7^{2})+7^{2}(1+7^{2})+7^{5}(1+7^{2})+7^{6}(1+7^{2})+..+7^{97}(1+7^{2})+7^{98}(1+7^{2})$
Dar observam ca
$1+7^{2}=1+49=50$ Atunci
$A=7*50+7^{2}*50+7^{5}*50+7^{6}*50+...+7^{97}*50+7^98*50=50(7+7^{2}+7^{5}+7^{6}+...+7^{97}+7^{98})$ care este un numar evident divizibil cu 50