Find all pairs of non-negative integers $x$ and one-digit natural numbers $y$ for which

$$
\frac{x}{y}+1=x, \overline{y}
$$

The notation on the right side of the equation denotes a repeating decimal.

(K. Pazourek)

Hint. Is there any connection between the decimal expansions of the numbers $\frac{x}{y}+1$ and $\frac{1}{y}$?
For $\frac{x}{y}+1$ to be a periodic number with a single-digit period, $\frac{x}{y}$ must also be a periodic number with a single-digit period. Since $\frac{x}{y}=x \cdot \frac{1}{y}$, the same condition must also apply to the number $\frac{1}{y}$. Among the natural numbers from 1 to 9, this condition is only met in three cases, which we will discuss in turn:

- For $y=3$, $\frac{1}{y}=0, \overline{3}$. Thus, the equation under discussion is of the form

$$
\frac{x}{3}+1=x+\frac{1}{3}
$$

which, after rearrangement, gives $x=1$.

- For $y=6$, $\frac{1}{y}=0,1 \overline{6}$. Thus, $0, \overline{6}=10 \cdot \frac{1}{6}-1=\frac{2}{3}$ and the equation under discussion is of the form

$$
\frac{x}{6}+1=x+\frac{2}{3},
$$

which, after rearrangement, gives $x=\frac{2}{5}$. However, this is not an integer.

- For $y=9$, $\frac{1}{y}=0, \overline{1}$. Thus, $0, \overline{9}=9 \cdot \frac{1}{9}=1$ and the equation under discussion is of the form

$$
\frac{x}{9}+1=x+1
$$

which, after rearrangement, gives $x=0$.

The problem has two solutions: $x=1, y=3$ and $x=0, y=9$.

Notes. Since $\frac{1}{9}=0, \overline{1}$, we have $\frac{y}{9}=0, \bar{y}$ and the equation under discussion can be expressed as

$$
\frac{x}{y}+1=x+\frac{y}{9}
$$

For each single-digit natural number $y$, it suffices to solve the corresponding linear equation and verify the non-negativity and integrality of $x$. This way, we obtain the two solutions mentioned above.

The above equation can be further rearranged, for example, as follows:

$$
\begin{aligned}
9 x+9 y & =9 x y+y^{2} \\
9(x+y-x y) & =y^{2}
\end{aligned}
$$

For any integers $x$ and $y$, the expression on the right-hand side is divisible by 9. Therefore, the number $y^{2}$ must also be divisible by 9, which means that the number $y$ must be divisible by 3. Among the numbers from 1 to 9, it suffices to verify only $y=3, 6$, and 9. This leads us to the same limitation of possibilities as in the approach described above.