## Problem 3

$3^{1}+3^{2}+3^{3}+\ldots+3^{2012}=\left(3^{1}+3^{2}+3^{3}+3^{4}\right)+3^{4}\left(3^{1}+3^{2}+3^{3}+3^{4}\right)+\ldots+3^{2008}\left(3^{1}+3^{2}+3^{3}+3^{4}\right)=$ (1p) $120+120 \cdot 3^{4}+\ldots 120 \cdot 3^{2008}=(\mathbf{1 p})$

$120\left(1+3^{4}+\ldots+3^{2008}\right)$ (1p)

It is sufficient to find the last digit of the number $1+3^{4}+\ldots+3^{2008}$ (1p)

The last digit of the numbers $1,3^{4}, \ldots, 3^{2008}$ is 1 (1p)

The last digit of the number $1+3^{4}+\ldots+3^{2008}$ is obtained by adding 1, 503 times, which is 3 (1p)

The last two digits of the number are 60 (1p)