Ta có :
\(A=1+2+2^2+.........+2^{2009}+2^{2010}\)
\(\Leftrightarrow A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+.......+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(\Leftrightarrow A=1\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...........+2^{2008}\left(1+2+2^2\right)\)
\(\Leftrightarrow A=1.7+2^3.7+.........+2^{2008}.7\)
\(\Leftrightarrow A=7\left(1+2^3+......+2^{2008}\right)⋮7\)
Vậy A chia 7 dư 0