Exercise 2. We take three digits $x, y, z$ such that $x>y>z>0$. By summing the six three-digit numbers obtained by permuting these 3 digits, we find 4884 (for example, if $x=3, y=2$ and $z=1$ we would have found $321+312+213+231+123+132=1332$). What are the possible values of the number formed by the three digits $x, y, z$ (in that order)? Justify your answer.
Solution to exercise 2 We have $(100 x+10 y+z)+(100 x+10 z+y)+(100 y+10 x+z)+(100 y+10 z+$ $x)+(100 z+10 x+y)+(100 z+10 y+x)=222(x+y+z)$, so $x+y+z=4884 / 222=22$.

If $x \leqslant 8$, then $y \leqslant 7$ and $z \leqslant 6$ so $x+y+z \leqslant 21$, which is impossible. Therefore, $x=9$ and $y+z=13$.

Since $y>z$, we have $y>13 / 2$ so $y=7$ or $y=8$. In the first case, we have $x=6$, and in the second case, we have $x=5$.

Conclusion: the solutions are 976 and 985.