Example 2 (Problem from the 3rd Canadian Mathematical Competition) 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.



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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 $x, y, z, u, v$ are one of the digits $0,1,2, \cdots, 8$ or 9, and $x \geqslant 1$.
$$
m=\overline{x y u v}=x \cdot 10^{3}+y \cdot 10^{2}+u \cdot 10+v \text {. }
$$

And $k=\frac{n}{m}$ is an integer. We can prove the inequality $9 m<n$, i.e.,
$9 \cdot\left(x \cdot 10^{3}+y \cdot 10^{2}+u \cdot 10+v\right)<x \cdot 10^{4}+y \cdot 10^{3}+z \cdot 10^{2}+u \cdot 10+v$, which is $80 u+8 v<10^{3} \cdot x+10^{2} \cdot y+10^{2} \cdot z$, and this is obviously true.
We can also prove the inequality $n<11 m$, i.e.,
$$
x \cdot 10^{4}+y \cdot 10^{3}+z \cdot 10^{2}+u \cdot 10+v<11 \cdot\left(x \cdot 10^{3}+y \cdot 10^{2}+u \cdot 10+v\right) \text {, }
$$

which is $10^{2} \cdot z<10^{3} x+10^{2} y+10^{2} u+10 v$, and this is obviously true.
Thus, $9 m<n<11 m$, i.e., $9<k<11$. Since $k$ is an integer, we have $k=10$.
Xian
Lin
Yuan
Ke
Mathematics
in
Therefore, $n=10 \mathrm{~m}$, i.e.,
$$
x \cdot 10^{4}+y \cdot 10^{3}+z \cdot 10^{2}+u \cdot 10+v=10\left(x \cdot 10^{3}+y \cdot 10^{2}+u \cdot 10+v\right) \text {, }
$$

which is $z \cdot 10^{2}=90 \cdot u+9 \cdot v=9(10 u+v)$.
Since 9 divides $z \cdot 10^{2}$, but 3 does not divide $10^{2}$, we have $z=9 t$ (where $t$ is an integer), thus $t \cdot 10^{2}=10 u+v$.
Clearly, $t=u=v=0$.
Therefore, we deduce that $n=\overline{x y 000}=N \cdot 10^{3}$, where $10 \leqslant N \leqslant 99$.