10-31 Let $n$ be a five-digit number (the first digit is not zero), and $m$ be the four-digit number formed by removing the middle digit of $n$. Determine all $n$ such that $\frac{n}{m}$ is an integer.
(3rd Canadian Mathematical Olympiad, 1971)
［Solution］Let $n=\overline{x y z u v}$
$$
=x \cdot 10^{4}+y \cdot 10^{3}+z \cdot 10^{2}+u \cdot 10+v .
$$

where $x, y, z, u, v \in\{0,1,2, \cdots, 9\}$, and $x \neq 0$.
By the problem statement,
$$
\begin{aligned}
m & =\overline{x y u v} \\
& =x \cdot 10^{3}+y \cdot 10^{2}+u \cdot 10+v .
\end{aligned}
$$

Let
$$
k=\frac{n}{m} .
$$

We seek $n$ such that $k$ is an integer.
Consider the difference between $n$ and $9 m$.
$$
\begin{aligned}
n-9 m= & \left(x \cdot 10^{4}+y \cdot 10^{3}+z \cdot 10^{2}+u \cdot 10+v\right) \\
& -\left(9 \cdot 10^{3} x+9 \cdot 10^{2} y+90 u+9 v\right) \\
= & 10^{3} x+10^{2} y+10^{2} z-80 u-8 v>0
\end{aligned}
$$

Thus,
$$
9 m0 .
\end{aligned}
$$

Thus,
$$
11 m>n \text {. }
$$

From (1) and (2),
$$
9 m<n<11 m
$$

Therefore,
$$
n=10 m, \quad k=10 . \\
x \cdot 10^{4}+y \cdot 10^{3}+z \cdot 10^{2}+u \cdot 10+v \\
=10\left(x \cdot 10^{3}+y \cdot 10^{2}+u \cdot 10+v\right) . \\
10^{2} z=90 u+9 v=9(10 u+v) .
$$

Since 9 is a divisor of $10^{2} z$, and 9 is coprime with $10^{2}$, 9 must be a divisor of $z$.
Let $z=9 t$, then
$$
t \cdot 10^{2}=10 u+v .
$$

Given $t, u, v \in\{0,1,2, \cdots, 9\}$, we have
$$
u=v=t=0 .
$$

Thus,
$$
z=9 t=0 \text {. }
$$

Therefore,
$$
n=\overline{x y 000} \text {. }
$$

Or, we can write $n=a \cdot 10^{3}$, where $10 \leqslant a \leqslant 99$.