Example 3 Add three digits after 764 to make the resulting six-digit number divisible by 8, 9, and 11.
Let the six-digit number be $\overline{764 a b c}$.
Let $M=7+4+b=11+b, N=c+a+6$.
Then (1) $\overline{a b c}$ is divisible by 8;
(2) $M+N=17+a+b+c$ is divisible by 9;
(3) $M-N=5+b-a-c$ is divisible by 11.

Now, estimate the range of $M+N$ and $M-N$.
By $17 \leqslant 17+a+b+c \leqslant 44$,
$-13 \leqslant 5+b-a-c \leqslant 14$,
we know
$$
\begin{array}{l}
M+N=18,27,36 ; \\
M-N=-11,0,11 .
\end{array}
$$

Enumerate in sequence:
When $M+N=18$, there is no solution;
When $M+N=27$, only $M-N=11$ has a solution, further solving gives $\overline{a b c}=82,181,280$, among which, only 280 is a multiple of 8;

When $M+N=36$, only $M-N=0$ has a solution, further solving gives $\overline{a b c}=478,676,874$, but none of them are multiples of 8.
In summary, the only solution to this problem is 764280.