Question

demonstrati ca numarul S= 1 + 7 + 7(la patrat) + 7(la cub) + ... 7(la puterea 2015) este divizibil cu 100

Answer 1

$S=(1+7+ 7^{2} + 7^{3})+( 7^{4} + 7^{5}+ 7^{6}+ 7^{7})+...+(7^{2012}+7^{2003}+ \\ +7^{2014}+7 ^{2015})= \\ = (1+7+7 ^{2}+7^{3})+7^{4}(1+7+7^{2} +7^{3})+...+7 ^{2012}(1+7+ 7^{2}+7^{3}) \\ =(1+7+7^{2}+7^{3})(1+7^{4}+7^{8}+...+7^{2012})= \\ =400 (1+7 ^{4}+ 7^{8}+...+7^{2012})= \\= \boxed{100} *4*(1+7^4+7^8+...+7^{2012}) \Rightarrow \boxed{S-divizibil~cu~100}.$