Câu hỏi của Nguyễn Minh Bảo Anh

Question

S=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 có chia hết cho (-6) không?

Nguyễn Huy Tú · Accepted Answer

$S=2+2^2+2^3+...+2^8$

$\Rightarrow S=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^7+2^8\right)$

$\Rightarrow S=6+2^2\left(2+2^2\right)+...+2^6\left(2+2^2\right)$

$\Rightarrow S=\left(1+2^2+...+2^6\right).6⋮6$

hay $S⋮\left(-6\right)$

Vậy...Câu trả lời: $S=2+2^2+2^3+...+2^8$

$\Rightarrow S=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^7+2^8\right)$

$\Rightarrow S=6+2^2\left(2+2^2\right)+...+2^6\left(2+2^2\right)$

$\Rightarrow S=\left(1+2^2+...+2^6\right).6⋮6$

hay $S⋮\left(-6\right)$

Vậy...

Violympic toán 6