1. Determine all natural numbers of the form $\overline{20 a 2 b 2}$ that are divisible by 72.
Solution. The number $\overline{20 a 2 b 2}$ is divisible by 72 if it is simultaneously divisible by 8 and by 9. To be divisible by 9, the sum of the digits must be divisible by 9, that is, $2+0+a+2+b+2=6+a+b$ must be divisible by 9.

At the same time, it must be divisible by 8, which means that the three-digit ending must be divisible by 8, that is, $\overline{2 b 2}$ must be divisible by 8. From this divisibility, it follows that $b \in\{3,7\}$. According to this, we conclude that for $b=3, a=0,9$; for $b=7, a=6$. Therefore, the required numbers are:

200232, 209232, and 205272.