Ta có :
\(A=1+2+2^2+....+2^{100}\)
\(\Leftrightarrow\left\{\begin{matrix}A=\left(1+2\right)+2\left(1+2\right)+.....+2^{99}\left(1+2\right)\\A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+....+2^{98}\left(1+2+2^2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}A=3+2.3+.....+2^{99}.3\\A=5+2^3.5+....+2^{98}.5\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}A⋮3\\A⋮5\end{matrix}\right.\)
Mà (3;5)=1
=> A chia hết cho 15
Vậy số dư của A khi chia cho 15 là 0
\(A=2^0+2^1+2^2+2^3+...+2^{100}\)
\(A=\left(2^0+2^1+2^2+2^3\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(A=1\left(2^0+2^1+2^2+2^3\right)+...+2^{57}\left(2^0+2^1+2^2+2^3\right)\)
\(A=\left(2^0+2^1+2^2+2^3\right)\left(1+...+2^{57}\right)\)
\(A=15.Q\)
\(\Rightarrow A⋮15\)
\(\Rightarrow A\div15\) dư \(0\)
Vậy \(A=2^0+2^1+2^2+2^3+...+2^{100}\div15\) dư \(0\)
