Ta có S=1+2+22+23+...+259
\(\Rightarrow\)2S=2+22+23+24+...+260
\(\Rightarrow\)2S-S=260-1
do 2 chia 3 dư 1 \(\Rightarrow\)260 chia 3 dư 160\(\Rightarrow\)260 chia 3 dư 1
\(\Rightarrow\)260 -1 \(⋮\)3
Hay S\(⋮\)3 (dpcm)
\(1+2+2^2+2^3+...+2^{59}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{58}+2^{59}\right)\)
\(=3+2^2\left(1+2\right)+...+2^{58}\left(1+2\right)\)
\(=3+2^2\times3+...+2^{58}\times3\)
\(=3\times\left(1+2^2+...+2^{58}\right)⋮3\)
Vậy \(S⋮3\)