Ta có: \(A=2+2^2+2^3+\cdots+2^{120}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\cdots+\left(2^{118}+2^{119}+2^{120}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+\cdots+2^{118}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+\cdots+2^{118}\right)\) ⋮7
Ta có: \(A=2+2^2+2^3+\cdots+2^{120}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+\cdots+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)
\(=2\left(1+2+\cdots+2^4\right)+2^6\left(1+2+\cdots+2^4\right)+\cdots+2^{116}\left(1+2+\cdots+2^4\right)\)
\(=31\left(2+2^6+\ldots+2^{116}\right)\) ⋮31
Ta có: \(A=2+2^2+2^3+\cdots+2^{120}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{119}+2^{120}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{119}\left(1+2\right)=3\left(2+2^3+\cdots+2^{119}\right)\) ⋮3
Ta có: A⋮3
A⋮7
mà ƯCLN(3;7)=1
nên A⋮3*7
=>A⋮21
