2. Determine the remainder of the division of the number $3^{100}$ by 13.
Solution. We have $3^{3}=2 \cdot 13+1$, i.e., $3^{3}=13 k+1$. Therefore,

$$
3^{3} \cdot 3^{3}=(13 k+1)^{2}=169 k^{2}+26 k+1=13\left(13 k^{2}+2\right)+1=13 m+1
$$

Then,

$$
\begin{aligned}
3^{3} \cdot 3^{3} \cdot 3^{3} & =(13 k+1)(13 m+1)=169 k m+13 k+13 m \\
& =13(13 k m+k+m)+1=13 n+1
\end{aligned}
$$

and so on. With further multiplication by $3^{3}$, we always get a number that gives a remainder of 1 when divided by 13. Therefore, $3^{99}=\left(3^{3}\right)^{33}=13 p+1$. Finally,

$$
3^{100}=3^{99} \cdot 3=3(13 p+1)=13 \cdot 3 p+3
$$

so the remainder when $3^{100}$ is divided by 13 is equal to 3.