When I divide the number $X$ by the number $Y$, I get the number $Z$ and a remainder of 27. When I divide the number $X$ by the number $Z$, I get the number $1.1 \cdot Y$ and a remainder of 3.

Which numbers $X, Y, Z$ satisfy the given conditions? Determine all possibilities.

(L. Hozová)
From the given, we have two equations:

$$
X=Y \cdot Z+27=1.1 \cdot Y \cdot Z+3
$$

By manipulating the second equation, we get $0.1 \cdot Y \cdot Z=24$, thus $Y \cdot Z=240$. Substituting back, we find that $X=267$.

Division with a remainder concerns integers. Therefore, all numbers $X, Y, Z$, and $1.1 \cdot Y$ must be integers. Hence, the number $Y$ must be a multiple of 10. The remainder is less than the divisor. Therefore, $Z \geq 4$ and $Y \geq 28$, which, together with the previous conclusion, gives $Y \geq 30$.

From the equation $Y \cdot Z=240$ and the requirement $Z \geq 4$, we get $Y \leq 240: 4=60$. Similarly, from the requirement $Y \geq 30$, we get $Z \leq 240: 30=8$. Thus, we have revealed two suitable pairs of numbers $Y$ and $Z$. All solutions can be obtained by systematically examining the possibilities within the given constraints:

| $Y$ | 30 | 40 | 50 | 60 |
| :---: | :---: | :---: | :---: | :---: |
| $Z$ | 8 | 6 |  | 4 |

The suitable triples of numbers $(X, Y, Z)$ are $(267,30,8)$, $(267,40,6)$, and $(267,60,4)$.

Evaluation. 1 point for expressing $Y \cdot Z=240$ and $X=267$; 1 point for each of the three solutions; 1 point for the quality of the commentary.

Note. In the given solution, it is advantageous to use the prime factorization $240=2^{4} \cdot 3 \cdot 5$.