I wrote a two-digit natural number on a card. The sum of the digits of this number is divisible by three. If I subtract the number 27 from the written number, I get another two-digit natural number, written with the same digits but in reverse order. Which numbers could I have written on the card?

(L. Hozová)
Let the digits of the two-digit number be denoted as follows: $x$ is in the tens place and $y$ is in the units place. The problem requires that the sum $x+y$ be divisible by three and

$$
10 x+y-27=10 y+x
$$

which simplifies to

$$
\begin{aligned}
9 x-9 y & =27, \\
x-y & =3 .
\end{aligned}
$$

Since the difference between the digits is three and their sum must be divisible by three, both digits $x$ and $y$ must be divisible by three. (This is evident if we express the sum of the digits $x+y$ as $y+3+y$.) The only possibilities are $x=9, y=6$ or $x=6, y=3$, i.e., the desired number can be 96 or 63. (The option 30 is not valid because 03 is not a two-digit natural number.)

Scoring. 1 point for writing the number and the number $\mathrm{s}$ with the digits in reverse order; 2 points for finding the relationship $x-y=3$; 2 points for determining one solution; 1 point for determining the second solution. (If the student lists 30 as a solution, it does not affect the scoring of their work.)

Alternative Solution. The desired number is greater than the number obtained by swapping the digits, so the digit in the units place of the desired number has a lower value than the digit in the tens place. Additionally, we know that the sum of these two digits is divisible by three. Therefore, we will go through all two-digit numbers whose digit sum is divisible by three and where the digit in the tens place is higher than the digit in the units place (the digit in the units place cannot be zero). For each, we find the number that is 27 less, and if it is written with the same digits, we have a solution.

| $96-27=69$ | 1st solution |
| :--- | :--- |
| $93-27=66$ |  |
| $87-27=60$ |  |
| $84-27=57$ |  |
| $81-27=54$ |  |
| $75-27=48$ |  |
| $72-27=45$ |  |
| $63-27=36$ | 2nd solution |
| $54-27=27$ |  |
| $51-27=24$ |  |
| $42-27=15$ |  |
| $21-27=-6$ |  |

The desired number can be 96 or 63.

Scoring. 3 points for explaining the principle of finding the numbers; 3 points for finding both solutions and ruling out the existence of other solutions.