Câu hỏi của linh angela nguyễn

Question

Cho . Số dư khi chia  cho 31 là

Phạm Nguyễn Tất Đạt · Accepted Answer

$A=1+2+2^2+2^3+...+2^{100}$(có 101 số)$A=1+\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)$$A=1+2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)$$A=1+2\cdot31+2^6\cdot31+...+2^{96}\cdot31$$A=1+31\left(2+2^6+...+2^{96}\right)$$\Rightarrow A:31$ dư 1Câu trả lời: $A=1+2+2^2+2^3+...+2^{100}$(có 101 số)$A=1+\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)$$A=1+2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)$$A=1+2\cdot31+2^6\cdot31+...+2^{96}\cdot31$$A=1+31\left(2+2^6+...+2^{96}\right)$$\Rightarrow A:31$ dư 1

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