2. The sum of a two-digit number and the two-digit number written with the same digits in reverse order, when divided by 5, gives a remainder of 2. What is that number?
Solution. Let $\overline{a b}$ be the two-digit number. We have

$$
\overline{a b}+\overline{b a}=10 a+b+10 b+a=10(a+b)+(a+b)
$$

Since $5 \mid 10$, the remainder when the sum $\overline{a b}+\overline{b a}$ is divided by 5 is equal to the remainder when the sum $a+b$ is divided by 5. Now, $a$ and $b$ are non-zero digits, so $a+b \leq 18$ and since the remainder when $a+b$ is divided by 5 must be equal to 2, the possible cases are:

- $a+b=2$ and since $a, b \neq 0$, we get $\overline{a b} \in\{11\}$,
- $\quad a+b=7$ and since $a, b \neq 0$, we get $\overline{a b} \in\{16,25,34,43,52,61\}$
- $\quad a+b=12$, so $\overline{a b} \in\{39,48,57,66,75,84,93\}$
- $a+b=17$, so $\overline{a b} \in\{89,98\}$.

Finally, $\overline{a b} \in\{11,16,25,34,39,43,48,52,57,61,66,75,84,89,93,98\}$.