12. Find the remainder of the division of $67^{283}$ by 13.
12. $65=13 \cdot 5$, then $67^{283}=(65+2)^{283}=65^{283}+283 \cdot 65^{282} \cdot 2+\ldots+2^{283}$. Each term, except the last one, contains a factor of 65, and therefore, is divisible by 13.

$$
2^{283}=16^{70} \cdot 2^{3}=(13+3)^{70} \cdot 2^{3}=\left(13^{70}+70 \cdot 13^{69} \cdot 3+\ldots+3^{70}\right) \cdot 2^{3}
$$

Each term, except the last one, is divisible by 13.

Consider the last term:

$$
3^{70} \cdot 2^{3}=27^{23} \cdot 3 \cdot 2^{3}=(26+1)^{23} \cdot 3 \cdot 2^{3}=\left(26^{23}+23 \cdot 26^{22}+\ldots+1\right) \cdot 3 \cdot 2^{3}
$$

Each term, except the last one, is divisible by 13, as it contains a factor of 26. Consider the last term.

$$
1 \cdot 3 \cdot 2^{3}=24=13+11 \text{. Answer: the remainder is } 11.
$$