\(M=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)=3.4+3^3.4+...+3^{99}.4=4\left(3+3^3+...+3^{99}\right)⋮4\)
\(M=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)=12+3^2.12+...+3^{98}.12=12\left(1+3^2+...+3^{98}\right)⋮12\)