1. Find all triples (not necessarily distinct) of digits $a, b, c$, for which the five-digit numbers $\overline{6 a b c 3}$ and $\overline{3 a b c 6}$ are in the ratio $63: 36$.
1. We will set up and solve the equation for the unknown digits $a, b, c$, which, due to the form of the given numbers, we can write directly for a single unknown $x=100a+10b+c$:

$$
\begin{aligned}
\frac{60000+10 x+3}{30000+10 x+6} & =\frac{63}{36}=\frac{7}{4}, \\
40 x+240012 & =70 x+210042, \\
30 x & =29970 \\
x & =999 .
\end{aligned}
$$

Conclusion. The found $x$ corresponds to the triplet of digits $a=b=c=9$. The problem has a unique solution.

For a complete solution, award 6 points, of which 3 points for setting up the correct equation and 3 points for solving it and the conclusion. For merely guessing the result (without explaining that it is the only possible one), award 1 point.