Let's take a three-digit number, for example, 346. By writing the digits in reverse order, we get 643. Subtracting the smaller number from the larger one $(643-346)$, we get 297. Writing the digits of this number in reverse order again: 792. The sum of these two numbers $297+792=1089$. Is it true that we always get 1089 as the result, no matter which three-digit number we choose?
I. solution. The general form of a 3-digit number is $100a + 10b + c$, where $a, b, c$ are single-digit numbers, including 0. Reversing the order of the digits gives: $100c + 10b + a$. We see that if $a = c$, the two numbers are equal, and their difference is zero. Therefore, we can assume that $a > c$, in which case the positive difference is $100(a - c) + (c - a)$. Since $c - a$ is negative, we break down one hundred into 9 tens and 1 ten, so that our number can be written with 3 non-negative digits according to the rules of writing numbers in the decimal system:

$$
100(a - c) + (c - a) = 100(a - c - 1) + 10 \cdot 9 + (10 + c - a)
$$

where $a - c - 1 = 0$ is also possible, in which case $10 + c - a = 10 - 1 = 9$.

Writing the digits of this number in reverse order:

$$
100(10 + c - a) + 10 \cdot 9 + (a - c - 1)
$$

The sum of these two numbers is:

$$
100(10 - 1) + 10 \cdot 18 + (10 - 1) = 900 + 180 + 9 = 1089
$$

Thus, we get 1089 as a result if and only if the digits in the units and hundreds places are not equal.

Pannonhalma g. circle.

II. solution. $(100a + 10b + c) - (100c + 10b + a) = 99(a - c)$, which is only non-zero if $a \neq c$. Let $a > c$. A 3-digit number divisible by 99 must have a middle digit of 9, as can be seen from the form $99x = 100x - x$, where $x$ represents a single-digit number. (In the case of $x = 1$, we consider 099 as a 3-digit number.) Due to divisibility by 9, the sum of the two outer digits must also be exactly 9. This implies divisibility by 11 as well $(9 - 9 = 0)$. Reversing the order of the digits, or swapping the two outer digits, does not affect divisibility by 9 and 11. Therefore, the new number is also divisible by 99, and thus the sum of the two numbers is a multiple of 99. If the value of the number before the swap was $99s$ (so the digit in the hundreds place was $s - 1$), then after the swap, the digit in the hundreds place will be $9 - (s - 1) = (10 - s)$, and thus the value of the new number will be $(11 - s)99$. The sum of the two numbers is therefore: $99s + 99(11 - s) = 11 \cdot 99 = 1089$.

György Pátkai (Bp. IX., Fáy András g. II. o. t.)