A four-digit palindrome is any four-digit natural number that has the same digit in the units place as in the thousands place and simultaneously has the same digit in the tens place as in the hundreds place. How many pairs of four-digit palindromes exist whose difference is 3674?

(L. Šimünek)
Let's denote the digits of the palindrome with letters. The fact that two four-digit palindromes have the required difference can be written as follows:

| $A B B A$ |
| ---: |
| $-C D D C$ |
| 3674 |

In the thousands and units columns, the same digits are subtracted. In the thousands column, we see that $A>C$. Therefore, when subtracting in the units column, there is no "carry over" and we have

$$
A-C=4 \text {. }
$$

In the tens and hundreds columns, where the same digits are subtracted, there is a "carry over" because in the result, the digit in the tens place is 1 greater than the digit in the hundreds place. We have

$$
10+B-D=7
$$

after rearranging this equation, we get

$$
D-B=3 \text {. }
$$

We will further analyze the thousands column to ensure that the given difference allows for the identification of suitable pairs of palindromes. Since there was a "carry over" when subtracting in the hundreds column, the digit in the thousands place in the result must be 1 less than the digit in the units place. We see that this is satisfied in the given difference.

In the problem, a four-digit palindrome is defined as a four-digit natural number with certain properties, so $A \neq 0$ and $C \neq 0$. To satisfy the relation (1), $A$ can only be equal to the digits $9, 8, 7, 6$ or 5, and $C$ is then always 4 less. From relation (2), we get that $D$ can only be equal to the digits $9, 8, 7, 6, 5, 4$ or 3, and $B$ is always 3 less. Thus, the letters $A$ and $C$ can be replaced by five different pairs of digits, and the letters $B$ and $D$ by seven different pairs of digits. These pairs can be combined in any way, giving us a total of $5 \cdot 7=35$ different pairs of palindromes with the desired properties.

Evaluation. 2 points for determining in which columns there is a carry over and in which there is not; 2 points for the counts of pairs $A, C$ and $B, D$; 2 points for the result. (Work in which $C$ is incorrectly substituted with 0 and which thus arrives at the result $6 \cdot 7=42$ can be rated up to 5 points.)