Ta có:
\(S=1+3+3^2+3^3+...+3^{100}\)
\(=\left(1+3+3^2+3^3+3^4\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=1\left(1+3+3^2+3^3+3^4\right)+...+3^{96}\left(1+3+3^2+3^3+3^4\right)\)
\(=1.121+...+3^{96}.121\)
\(=121\left(1+...+3^{96}\right)⋮121\)
Vậy \(S\div121\) có chữ số tận cùng là \(0\)