3. Three four-digit numbers $k, l, m$ have the same form $A B A B$ (i.e., the digit in the units place is the same as the digit in the hundreds place, and the digit in the tens place is the same as the digit in the thousands place). The number $l$ has the digit in the units place 2 greater and the digit in the tens place 1 less than the number $k$. The number $m$ is the sum of the numbers $k$ and $l$ and is divisible by nine. Determine all such numbers $k$.
SOLUTION. For the number $m=\overline{C D C D}$ to be divisible by nine, the sum $2(C+D)$ of its digits must be divisible by nine, and thus the sum $C+D$ must also be divisible by nine, meaning the number $\overline{C D}$ must be divisible by nine.

If the number $k$ has digits $A, B, A, B$, and the number $l$ has digits $A-1, B+2, A-1, B+2$, since the number $B+(B+2)=2 B+2$ is even, the digit $D$ of the number $m=k+l$ must be even. Therefore, in terms of divisibility by nine, only the following numbers $m$ are possible: 1818, 3636, 5454, 7272, 9090. Since the digit $C$ is odd in all cases and the sum of the digits $A+(A-1)=2 A-1$ is also odd, $B+(B+2)$ cannot be greater than 10, so $B+(B+2)=D$ and $A+(A-1)=C$. From this, we can easily determine the corresponding digits $C, D$ and write the numbers $k, l$ in the following table:

| $m$ | 1818 | 3636 | 5454 | 7272 | 9090 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $k$ | 1313 | 2222 | 3131 | 4040 | does not exist |
| $l$ | 0505 | 1414 | 2323 | 3232 | does not exist |

The number 0505 is not a four-digit number, so the solutions to the problem are only the numbers $k \in\{2222,3131, 4040\}$

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