\(A=2+2^2+2^3+2^4+...+2^9+2^{10}\\ A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\\ A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^8\left(2+2^2\right)\\ A=\left(2+2^2\right)\left(1+2^2+...+2^8\right)\\ A=6\cdot\left(1+2^2+...+2^8\right)⋮3\)
________________________________________________
\(A=2+2^2+2^3+2^4+...+2^9+2^{10}\\ A=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\\ A=\left(2+2^2+2^3+2^4+2^5\right)+2^5\left(2+2^3+2^4+2^5\right)\\ A=\left(2+2^2+2^3+2^4+2^5\right)\left(1+2^5\right)\\ A=62\cdot33\)
Ta thấy \(\left\{{}\begin{matrix}62\equiv6\left(mod7\right)\\33\equiv5\left(mod7\right)\end{matrix}\right.\)
\(\Rightarrow A\equiv2\left(mod7\right)\)
hay A chia 7 dư 2
