Ta có:
\(A=2+2^2+2^3+2^4+....+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+.....+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+....+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(=2\times31+....+2^{96}\times31\)
\(=31\left(2+2^6+....+2^{96}\right)\)
\(\Rightarrow\)A chia hết cho 31 (vì có chứa thừa số 31)
Nhưng 2 + 2^2 + 2^3 + 2^4 + 2^4 + 2^5 = 62 ko ra 31