Task B-2.4. Determine all seven-digit numbers of the form $\overline{2012 x y x}$ that are divisible by 72.
Solution. A number is divisible by 72 if it is divisible by 8 and if it is divisible by 9.

For a number to be divisible by 9, the sum of its digits must be divisible by 9.

For a number to be divisible by 8, its three-digit ending, the number $\overline{x y x}$, must be divisible by 8.

The sum of the digits of the given number is $5+2 x+y$.

Since $x, y \leq 9$, we have

$$
\begin{aligned}
& 5+2 x+y=9 \text{ or } \\
& 5+2 x+y=18 \text{ or } \\
& 5+2 x+y=27
\end{aligned}
$$

(1) Let $5+2 x+y=9$. Then $y=4-2 x$.

The number $\overline{x y x}$ can be written as

$$
\overline{x y x}=100 x+10 y+x=101 x+10(4-2 x)=81 x+40=80 x+40+x.
$$

This will be divisible by 8 only if $x=8$ or $x=0$. However, for $x=8$, $y=4-2 \cdot 8=-12$, which is impossible. If $x=0$, then $y=4$, so the desired number is 2012040.

(2) Let $5+2 x+y=18$. Then $y=13-2 x$.

The number $\overline{x y x}$ can be written as

$$
\overline{x y x}=101 x+10 y=81 x+130=80 x+128+2+x.
$$

This will be divisible by 8 only if $2+x=8$, that is, $x=6$. Then $y=13-2 \cdot 6=1$.

(3) Let $5+2 x+y=27$. Then $y=22-2 x$.

The number $\overline{x y x}$ can be written as

$$
\overline{x y x}=101 x+10 y=81 x+220=80 x+216+4+x
$$

This will be divisible by 8 only if $4+x=8$, that is, $x=4$. Then $y=22-2 \cdot 4=14$, which is impossible.

Therefore, the only solutions are the numbers 2012616 and 2012040.