Câu hỏi của Tô Hồng Đức

Question

Cho N = 2^5 + 2^9 + 2^13 + ... + 2^409

a.CMR:N chia hết cho 17

b.CMR:(2^413 - 2^5) chia hết cho 17

Quang Ho Si · Accepted Answer

a, $N=2^5\left(1+2^4\right)+2^{13}\left(1+2^4\right)+.....+2^{405}\left(1+2^4\right)$

$\Rightarrow N=\left(1+2^4\right)\left(2^5+2^{13}+...+2^{405}\right)=17\left(2^5+2^{13}+...+2^{405}\right)$

vậy N chia hết cho 17Câu trả lời: a, $N=2^5\left(1+2^4\right)+2^{13}\left(1+2^4\right)+.....+2^{405}\left(1+2^4\right)$

$\Rightarrow N=\left(1+2^4\right)\left(2^5+2^{13}+...+2^{405}\right)=17\left(2^5+2^{13}+...+2^{405}\right)$

vậy N chia hết cho 17