Câu hỏi của Bùi Thanh Tâm

Question

Tìm số dư của phép chia A=$2^0+2^1+2^2+2^3+..........+2^{2015}+2^{2016}$ cho 7

Nguyễn Việt Lâm · Accepted Answer

$A=1+\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)$

$=1+2\left(1+2+4\right)+2^4\left(1+2+4\right)+...+2^{2014}\left(1+2+4\right)$

$=1+2.7+2^4.7+...+2^{2014}.7$

$=1+7\left(2+2^4+...+2^{2014}\right)$

$\Rightarrow A$ chia 7 dư 1Câu trả lời: $A=1+\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)$

$=1+2\left(1+2+4\right)+2^4\left(1+2+4\right)+...+2^{2014}\left(1+2+4\right)$

$=1+2.7+2^4.7+...+2^{2014}.7$

$=1+7\left(2+2^4+...+2^{2014}\right)$

$\Rightarrow A$ chia 7 dư 1

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