$10 \cdot 32$ In the number 3000003, what digits should the hundred's place digit 0 and the ten-thousand's place digit 0 be replaced with, so that the resulting number is divisible by 13?
(Polish Mathematical Competition, 1950)
[Solution] Let the required digits be $x$ and $y$, where $x, y$ satisfy $0 \leqslant x \leqslant 9, 0 \leqslant y \leqslant 9, \quad x, y \in Z$.
Let $m=30 x 0 y 03$, then
$$
m=3 \cdot 10^{6}+x \cdot 10^{4}+y \cdot 10^{2}+3 \text {. }
$$

Notice that
$$
10^{6} \equiv 1(\bmod 13) \text {, }
$$
$$
\begin{array}{l}
10^{4} \equiv 3 \quad(\bmod 13), \\
10^{2} \equiv 9 \quad(\bmod 13),
\end{array}
$$

Therefore,
$$
\begin{aligned}
m & =3 \cdot 10^{6}+x \cdot 10^{4}+y \cdot 10^{2}+3 \\
& \equiv 3 x+9 y+6 \quad(\bmod 13) \\
& \equiv 3(x+3 y+2) \quad(\bmod 13) .
\end{aligned}
$$

Since $(3,13)=1$, for $13 \mid m$ to hold,
$$
13 \mid x+3 y+2 \text {. }
$$

Given $0 \leqslant x \leqslant 9, 0 \leqslant y \leqslant 9$, we have
$$
2 \leqslant x+3 y+2 \leqslant 38 \text {. }
$$

Thus, $x+3 y+2=13$ or $x+3 y+2=26$.
(1) When

Thus,
$$
\begin{array}{rl}
x+3 y+2 & =13 \\
x & \text {, } \\
x & 11-3 y \geqslant 0, \\
y \leqslant \frac{11}{3}, \quad y \in\{0,1,2,3\} .
\end{array}
$$

Clearly, $y \neq 0$, so there are three integer solutions:
$$
\left\{\begin{array} { l } 
{ x = 2 , } \\
{ y = 3 , }
\end{array} \quad \left\{\begin{array} { l } 
{ x = 5 , } \\
{ y = 2 , }
\end{array} \left\{\begin{array}{l}
x=8, \\
y=1 .
\end{array}\right.\right.\right.
$$
(2) When $x+3 y+2=26$,
$$
0 \leqslant x=3(8-y) \leqslant 9 \text {, }
$$

Thus,
$$
5 \leqslant y \leqslant 8
$$

Therefore, there are four integer solutions:
$$
\left\{\begin{array} { l } 
{ x = 9 , } \\
{ y = 5 , }
\end{array} \quad \left\{\begin{array} { l } 
{ x = 6 , } \\
{ y = 6 , }
\end{array} \quad \left\{\begin{array} { l } 
{ x = 3 , } \\
{ y = 7 , }
\end{array} \quad \left\{\begin{array}{l}
x=0, \\
y=8 .
\end{array}\right.\right.\right.\right.
$$

From (1) and (2), we get 7 integer solutions:
$$
\begin{array}{l}
3080103,3050203,3020303,3090503, \\
3060603,3030703,3000803 .
\end{array}
$$