Problem 1. Determine the natural numbers of the form $\overline{ab}$, which have the property that the remainder of the division of $\overline{ab}$ by $a+b$ is $a \cdot b$.
Solution. We have the condition $a \cdot b < a + b$. If $a$ and $b$ were both greater than or equal to 2, assuming $a \geq b$, we get $a \cdot b \geq 2 \cdot a \geq a + b$, a contradiction. The same applies for $a \leq b$. Therefore, $a \leq 1$ or $b \leq 1$.

Case I. For $a \leq 1$, since $a \neq 0$, we get $a = 1$. According to the division theorem, we have $\overline{1 b} = (1 + b) \cdot C + b$, from which $10 = (1 + b) \cdot C$, where $C$ is the quotient of the division. Thus, $b + 1$ divides 10, and $b$ can take the values $0, 1, 4$ or 9. We obtain the solutions $10, 11, 14, 19 \ldots \ldots .2$ p

Case II. For $b \leq 1$, we have two subcases:

a) If $b = 0$, we obtain the solutions $10, 20, \ldots, 90$.

b) If $b = 1$, from the division theorem, we get $\overline{a 1} = (a + 1) \cdot C + a$, from which $9 \cdot (a + 1) = (a + 1) \cdot C + 8$. We obtain that $a + 1$ divides 8, so $a$ can be 1, 3, or 7. We get the solutions $11, 31, 71$ $.2 p$