Example 24 (Problem from the 48th Slovenian Mathematical Olympiad) Find all five-digit numbers $\overline{a b c d e}$, such that the number is divisible by 9, and $\overline{a c e}-\overline{b d a}=760$.
The number theory problem to solve the equation $\overline{a c e}-\overline{b \overline{d a}}=760$ can be rewritten as
$100 a+10 c+e-100 b-10 d-a=760$,
from which we can deduce that $e=a$.
Dividing both sides of the equation by 10, we get
$10(a-b)+(c-d)=76$.
Therefore, there are only two possibilities:
$c-d=6$ or $c-d=-4$.
If $c-d=6$, then we have
$c=d+6, a=b+7$.
Since the five-digit number is divisible by 9, we have
$$
\begin{aligned}
a+b+c+d+e & =b+7+b+d+6+d+b+7 \\
& =3 b+2 d+20=3 b+2(d+1+9)
\end{aligned}
$$

which must be divisible by 9.
So, $d+1$ must be divisible by 3.
Since $c-d=6$, we get $d=2$, thus $c=8$.
At the same time, $3(b+2)$ must be divisible by 9.
Since $a=b+7$, we have $b=1$.
In this case, the desired five-digit number is 81828.
If $c-d=-4$, then we have
$$
d=c+4, a=b+8 \text {. }
$$

Therefore, $a=8, b=0$, or $a=9, b=1$.
If $a=8$, by
$$
9 \mid(a+b+c+d+e)=8+2 c+4+8,
$$

we can deduce that $2 c+2$ must be divisible by 9.
Thus, $c=8$, and consequently $d=c+4=12$, which is not a single digit.
If $a=9$, by
$$
9 \mid(a+b+c+d+e)=10+2 c+4+9,
$$

we can deduce that $2 c+5$ must be divisible by 9.
Thus, $c=2$, and the corresponding five-digit number is 91269.