1. A four-digit number has the digits in the thousands and tens places as 7 and 2, respectively. Which digits should be placed in the hundreds and units places so that the number is divisible by 2 and 3, but not by their higher powers?
Solution. If the hundreds digit is $a$, and the units digit is $b$, then the number is of the form $\overline{7 a 2 b}, a, b \in\{0,1,2, \ldots, 9\}$. For the number $\overline{7 a 2 b}$ to be divisible by 2, $b \in\{0,2,4,6,8\}$, but since the number should not be divisible by higher powers of the number 2, we conclude that its two-digit ending should not be divisible by 4, so $b \in\{2,6\}$.

a) If $b=2$, then the number is of the form $\overline{7 a 22}$ and since it must be divisible by 3, the sum of the digits $7+a+2+2=11+a$ must be divisible by 3, so $a \in\{1,4,7\}$. However, for $a=7$ we get the number 7722 which is divisible by $9=3^{2}$, which contradicts the condition of the problem. Therefore, in this case, the numbers 7122 and 7422 are solutions to the problem.

b) If $b=6$, then the number is of the form $\overline{7 a 26}$ and since it must be divisible by 3, the sum of the digits $7+a+2+6=15+a$ must be divisible by 3, so $a \in\{0,3,6,9\}$. However, for $a=3$ we get the number 7326 which is divisible by $9=3^{2}$, which contradicts the condition of the problem. Therefore, in this case, the numbers 7026, 7626, and 7926 are solutions to the problem.

Finally, the solutions to the problem are the numbers 7026, 7122, 7422, 7626, and 7926.