Problem 859

1. We will gradually transform the given equality for $z \neq 0$:

$$\begin{aligned} \frac{x+y}{z} & =\overline{z, y x}, \\ \frac{x+y}{z} & =z+\frac{y}{10}+\frac{x}{100}, \\ 100(x+y) & =(100 z+10 y+x) \cdot z . \end{aligned}$$

Since $x, y, z$ are digits, the inequalities $100 \cdot(9+9) \geq 100(x+y)$ and $(100 z+10 y+x) \cdot z \geq 100 z \cdot z$ hold, from which it follows that $18 \geq z^{2}$. This means that $z$ can only be $1$, $2$, $3$, or $4$ (the value $z=0$ is not permissible).

For $z=1$, the given equation takes the form

$$\begin{aligned} & 100(x+y)=100+10 y+x, \\ & 99 x+90 y=100 . \end{aligned}$$

By considering divisibility by three or nine, we find that the last equation has no integer solutions. Therefore, $z$ cannot be $1$.

For $z=2$, the given equation takes the form

$$\begin{aligned} 100(x+y) & =(200+10 y+x) \cdot 2, \\ 49 x+40 y & =200 . \end{aligned}$$

By considering divisibility by ten, we find that $x$ can only be $0$. Then $y=5$, so in this case, the digits $x=0, y=5, z=2$ satisfy the given equation.

For $z=3$, the given equation takes the form

$$\begin{aligned} 100(x+y) & =(300+10 y+x) \cdot 3, \\ 97 x+70 y & =900 . \end{aligned}$$

By considering divisibility by ten, we find that $x$ can only be $0$. However, there is no integer $y$ that satisfies the equation $70 y=900$. Therefore, $z$ cannot be $3$.

For $z=4$, the given equation takes the form

$$\begin{aligned} 100(x+y) & =(400+10 y+x) \cdot 4, \\ 24 x+15 y & =400 . \end{aligned}$$

By considering divisibility by three, we find that the last equation has no integer solutions. Therefore, $z$ cannot be $4$.

The given equation is satisfied only for $x=0, y=5, z=2$. Indeed, $\frac{0+5}{2}=2.50$.

For a complete solution, award 6 points. If the solution is carried out in the manner described, award 2 points for the restriction $z<5$, and 1 point for solving the equation for each individual value of $z \in\{1,2,3,4\}$.