A=\(2^0+2^1+2^2+2^3+2^4+...+2^{2016}\)
\(=\left(2^0+2^1+2^2\right)+\left(2^3+2^4+2^5+......+2^{2016}\right)\)
=7+\(2^3\left(1+2^1+2^2+....+2^{2012}+2^{2013}\right)\)
\(=7+8\left(1+2+...+2^{2013}\right)\)
Vì \(8\left(1+2+...+2^{2012}+2^{2013}\right)⋮8\)
Mà \(7⋮̸8\) Nên \(7+8\left(1+2+..+2^{2012}+2^{2013}\right)\div8\) sẽ dư 7
Hay \(A\div8\) dư 7