Let $a$ and $b$ be two non-zero digits not necessarily different. The two-digit number $\overline{a b}$ is called curious if it is a divisor of the number $\overline{b a}$, which is formed by swapping the order of the digits of $\overline{a b}$. Find all curious numbers.

Note: The bar over the numbers serves to distinguish the product $a \cdot b$ from the two-digit number $\overline{a b}$.
Solution

The two-digit number $\overline{a b}$ can be written as $10 a+b$, and similarly, $\overline{b a}=10 b+a$. If $10 a+b$ is a divisor of $10 b+a$, we have $10 b+a=(10 a+b) k$, where $k$ is an integer less than or equal to 9 since both numbers have two digits. It follows that

$$
\begin{aligned}
10 b+a & =(10 a+b) k \\
10 b+a+10 a+b & =(10 a+b)(k+1) \\
11(a+b) & =(10 a+b)(k+1)
\end{aligned}
$$

From the last equation, the number $(10 a+b)(k+1)$ must be a multiple of 11. Since $k \leq 9$, we have $k+1 \leq 10$ and, consequently, $k+1$ does not have a factor of 11, implying that $10 a+b$ must be a multiple of 11. All two-digit multiples of 11 have equal digits $\mathrm{and}$ this allows us to conclude that necessarily $a=b$. When $a=b, \overline{a b}=\overline{b a}$ and, certainly, one divides the other. Therefore, the set of curious numbers is

$\{11,22,33,44,55,66,77,88,99\}$.