Example 1 Given that $38 a 75 b$ is a multiple of 11. Find $a, b$.

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Solve: By rule 3, we get
$$
b+7+8-(5+a+3)=7+b-a
$$

is a multiple of 11. Let $7+b-a=11k$. Then
$11k=7+b-a \leqslant 7+b \leqslant 16$
$\Rightarrow k \leqslant 1$.
Also, $11k=7+b-a \geqslant 7+0-9=-2$, so $k \geqslant 0$.
Thus, $0 \leqslant k \leqslant 1$.
(1) When $k=0$, $a-b=7$, at this time, $(a, b)=(7,0),(8,1),(9,2)$.
(2) When $k=1$, $b-a=4$, at this time,
$$
\begin{aligned}
(a, b)= & (0,4),(1,5),(2,6),(3,7), \\
& (4,8),(5,9) .
\end{aligned}
$$

Upon verification, all nine solutions satisfy the condition.