Ta có: 155 = 5.31 ta chứng minh A chia hết cho 5 và 31
+ Chứng minh A chia hết cho 5
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+4+8\right)+2^5\left(1+2+4+8\right)+...+2^{97}\left(1+2+4+8\right)\)
\(=15\left(2+2^5+...+2^{97}\right)=3.5.\left(2+2^5+...+2^{97}\right)\)
\(\Rightarrow A⋮5\left(1\right)\)
+ Chứng minh A chia hết cho 31
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+4+8+16\right)+2^6\left(1+2+4+8+16\right)+...+2^{96}\left(1+2+4+8+16\right)\)
\(=31\left(2+2^6+...+2^{96}\right)\)
\(\Rightarrow A⋮31\left(2\right)\)
Từ (1) và (2) \(\Rightarrow A⋮\left(31.5\right)hayA⋮155\)