## Task 2 - 120712

Determine all three-digit natural numbers $z$ that each satisfy the following conditions simultaneously:

(1) The number $z$ is divisible by both 9 and 11.

(2) If the digit in the hundreds place of the number $z$ is swapped with the digit in the units place, a new three-digit number $z^{\prime}$ is obtained, which is $\frac{2}{9}$ of the number $z$.
Assume a three-digit natural number $z$ has the properties (1) and (2).

Due to (1) and because 9 and 11 are coprime, $z$ is a multiple of 99. Since $z$ is a three-digit number, the possible numbers are

$$
198,297,396,495,594,693,792,891,990
$$

that can satisfy conditions (1) and (2).

The number 990 is excluded because swapping the units digit with the hundreds digit does not result in a three-digit number. Furthermore, since $z'$ must be less than $z$, the numbers 198, 297, 396, and 495 are also excluded. Finally, since $z' = \frac{2}{9} z$ must be an even number, the hundreds digit of $z$ must be even, so the numbers 594 and 792 are also excluded. For the remaining two numbers, we get:

| $z$ | $\frac{2}{9} z$ | $z'$ |
| :---: | :---: | :---: |
| 693 | 154 | 396 |
| 891 | 198 | 198 |

Thus, only the number $z=891$ can satisfy conditions (1) and (2). A check confirms that it indeed has the properties (1) and (2).