$\qquad$Example 12 (2003 Japan Mathematical Olympiad) Given two distinct positive integers $a$ and $b$, and $b$ is a multiple of $a$. If represented in decimal, both $a$ and $b$ consist of $2n$ digits, and the most significant digit is non-zero. Also, the first $n$ digits of $a$ and the last $n$ digits of $b$ are the same, and vice versa, for example, $n=2, a=1234, b=3412$ (but this example does not satisfy the condition that $b$ is a multiple of $a$), find $a$ and $b$.

$\frac{\operatorname{\pi }}{\pi}$
Let $a$ and $b$ be represented by two $n$-digit positive integers $x$ and $y$, then
$$
a=10^{n} x+y, b=10^{n} y+x \text {. }
$$

Given that $x<y$, and $10^{n} y+x=m\left(10^{n} x+y\right)$, where $2 \leqslant m \leqslant 9$.
Adding $10^{n} x+y$ to both sides, we get
$$
(x+y)\left(10^{n}+1\right)=(m+1)\left(10^{n} x+y\right) \text {. }
$$

Assume that $m+1$ and $10^{n}+1$ are coprime, then
$$
10^{n} x+y \equiv 0\left(\bmod \left(10^{n}+1\right)\right) \text {. }
$$

Thus, $x \equiv y\left(\bmod \left(10^{n}+1\right)\right)$.
Since $x, y$ are only $n$ digits, this contradicts $x<y$.
Since $m+1 \leqslant 10$, $m+1$ and $10^{n}+1$ must have a common prime factor, which must be one of the integers 2, 3, 5, 7. However, 2, 3, 5 cannot divide $10^{n}+1$, so it must be 7. Therefore, $m=6$, and $7 k=10^{n}+1$.
Thus, equation (1) becomes
$(x+y) 7 k=7[(7 k-1) x+y]$, which simplifies to $5 k x=(k-1)(y-x)$.
Noting that $7(k-1)=10^{n}-6$, we see that 5 and $k-1$ are coprime.
Therefore, $5 k$ and $k-1$ are also coprime.
Thus, $y-x$ is a multiple of $5 k$.
Since $0<y-x<10^{n}+1=7 k$, then $y-x=5 k$.
Therefore, $x=k-1, y=6 k-1$, and
$$
\begin{aligned}
a & =10^{n} x+y=10^{n}(k-1)+(6 k-1) \\
& =10^{n}\left[\frac{1}{7}\left(10^{n}+1\right)-1\right]+\left[\frac{6}{7}\left(10^{n}+1\right)-1\right]=\frac{1}{7}\left(10^{2 n}-1\right), \\
b & =m a=\frac{6}{7}\left(10^{2 n}-1\right) .
\end{aligned}
$$

where, $n \equiv 3(\bmod 6)$ (since $10^{n}+1$ is a multiple of 7).