2. The product of two two-digit numbers is 4032. The second number is written with the same digits as the first, but in reverse order. What are these numbers?
Solution. Let's denote the given two-digit numbers as $\overline{\mathrm{ab}}$ and $\overline{\mathrm{ba}}$. Then, from the condition in the problem, we have $\overline{\mathrm{ab}} \cdot \overline{\mathrm{ba}}=4032=2^{6} \cdot 3^{2} \cdot 7$. From this, it follows that 3 is a divisor of both numbers $\overline{a b}$ and $\overline{\mathrm{ba}}$ (since if one of them is divisible by 9, then the other will also be divisible by 9, meaning the product of the two numbers is divisible by 81, which is incorrect). Furthermore, 7 is a divisor of only one of the numbers, for example, $\overline{\mathrm{ab}}$. This means 21 is a divisor of $\overline{\mathrm{ab}}$, i.e., $\overline{\mathrm{ab}} \in\{21,42,84\}$. By checking, it is found that the required numbers are 84 and 48.