4. Determine the digits $a$ and $b$ such that $13 \mid \overline{2 a 0 b 82}$.
Solution. We have

$$
\begin{aligned}
\overline{2 a 0 b 82} & =200082+10000 a+100 b \\
& =(15390 \cdot 13+12)+(769 \cdot 13+3) a+(7 \cdot 13+9) b \\
& =13 \cdot 16166+(12+3 a+9 b) .
\end{aligned}
$$

Thus, the digits $a$ and $b$ need to be determined from the condition $13 \mid 12+3 a+9 b$. We have,

- if $b=0$, then $12+3 a+9 b=12+3 a=3(4+a)$ and since $13 \mid 3(4+a)$, we get $a=9$,
- if $b=1$, then $12+3 a+9 b=21+3 a=3(7+a)$ and since $13 \mid 3(7+a)$, we get $a=6$,
- if $b=2$, then $12+3 a+9 b=30+3 a=3(10+a)$ and from $13 \mid 3(10+a)$, we get $a=3$,
- if $b=3$, then $12+3 a+9 b=39+3 a=3(13+a)$ and from $13 \mid 3(13+a)$, we get $a=0$,
- if $b=4$, then $12+3 a+9 b=48+3 a=3(16+a)$ and from $13 \mid 3(16+a)$, the problem has no solution in this case,
- if $b=5$, then $12+3 a+9 b=57+3 a=3(19+a)$ and from $13 \mid 3(19+a)$, we get $a=7$,
- if $b=6$, then $12+3 a+9 b=66+3 a=3(22+a)$ and from $13 \mid 3(22+a)$, we get $a=4$,
- if $b=7$, then $12+3 a+9 b=75+3 a=3(25+a)$ and from $13 \mid 3(25+a)$, we get $a=1$,
- if $b=8$, then $12+3 a+9 b=84+3 a=3(28+a)$ and from $13 \mid 3(28+a)$, the problem has no solution in this case,
- if $b=9$, then $12+3 a+9 b=93+3 a=3(31+a)$ and from $13 \mid 3(31+a)$, we get $a=8$.