## Task B-2.6.

In the process of simplifying the fraction $\frac{\overline{200 \ldots 0 x}}{\overline{300 \ldots 0 y}}(x, y \neq 0)$, where there are 2020 zeros between 2 and $x$ and between 3 and $y$, Matko neglected the zeros and wrote $\frac{\overline{200 \ldots 0 x}}{\overline{300 \ldots 0 y}}=\frac{\overline{2 x}}{\overline{3 y}}$ and obtained the correct result. Determine all values of the digits $x$ and $y$ for which Matko could have obtained the correct result with the described simplification.
## Solution.

From the equation $\frac{\overline{200 \ldots 0 x}}{\overline{300 \ldots 0 y}}=\frac{\overline{2 x}}{\overline{3 y}}$, we get the equation $\frac{2 \cdot 10^{2021}+x}{3 \cdot 10^{2021}+y}=\frac{20+x}{30+y}$.

After rearranging, we have

$$
\begin{array}{cc}
6 \cdot 10^{2022}+20 y+3 x \cdot 10^{2021}+x y=6 \cdot 10^{2022}+30 x+2 y \cdot 10^{2021}+x y, & 1 \text { point } \\
20 y+3 x \cdot 10^{2021}=30 x+2 y \cdot 10^{2021}, & \\
10^{2021}(3 x-2 y)=10(3 x-2 y), & 1 \text { point } \\
\left(10^{2021}-10\right)(3 x-2 y)=0 . & 1 \text { point } \\
1 \text { point }
\end{array}
$$

This equality will only hold if $3 x-2 y=0$ or $3 x=2 y$. 2 points

Since $x$ and $y$ are digits different from 0, the solutions are $(x, y) \in\{(2,3),(4,6),(6,9)\}$. 2 points