Question

Arata ca: 1. a) (9^1+9^2+...+9^100) divizibil cu 100
B) (7•2^11•9^10+5•2^10•3^21-4•2^10•9^10) divizibil cu 2

Answer 1

$b)(7\cdot2^{11}\cdot9^{10}+5\cdot2^{10}\cdot3^{21}-4\cdot2^{10}\cdot9^{10})= \\ \\ (7\cdot2^{11}\cdot3^{20}+5\cdot2^{10}\cdot3^{21}-4\cdot2^{10}\cdot3^{20})= \\ \\ 2^{10}\cdot3^{20}(7\cdot2+5\cdot3-4)= \\ \\ 2^{10}\cdot3^{20}(14+15-4)= \\ \\ 2^{10}\cdot3^{20}\cdot25= \\ \\ 2^{10}\cdot3^{20}\cdot 5^{2} =$

deci este vizibil divizibil cu $2^{10}$  deci este vizibil divizibil cu 2