10・11 If $a, b, c, d, e, f, p, q$ are Arabic numerals, and $b>c>d>a$. The difference between the four-digit numbers $\overline{c d a b}$ and $\overline{a b c d}$ is a four-digit number of the form $\overline{p q e f}$. If $\overline{e f}$ is a perfect square, and $\overline{p q}$ is not divisible by 5. Find the four-digit number $\overline{a b c d}$, and briefly explain the reasoning.
(China Beijing Junior High School Grade 3 Mathematics Competition, 1983)
[Solution]

That is,
$$
\begin{array}{l} 
\overline{\text { pqef }} \\
= \overline{c d a b}-\overline{a b c d} \\
=\left(\overline{c d} \cdot 10^{2}+\overline{a b}\right)-\left(\overline{a b} \cdot 10^{2}+\overline{c d}\right) \\
= 99(\overline{c d}-\overline{a b}),
\end{array}
$$
$$
\begin{array}{c}
\overline{p q} \cdot 100+\overline{e f}=99(\overline{c d}-\overline{a b}) . \\
99 \cdot \overline{p q}+(\overline{p q}+\overline{e f})=99(\overline{c d}-\overline{a b}) .
\end{array}
$$

Thus,
$$
99 \mid(\overline{p q}+\overline{e f}) \text {. }
$$

But since $\overline{e f}$ is a perfect square,

then
thus
$$
\begin{array}{c}
\overline{\overline{e f}} \leqslant 81, \\
\overline{p q}+\overline{e f}c>d .
$$
$$
f=b-d \geqslant 2 \text {. }
$$

So the unit digit
Therefore, the unit digit of the perfect square $\overline{e f}$ can only be 5 or 6.
Note the subtraction in the vertical form

In which, because $b>c>d>a$, we know $c \leqslant 8, a \geqslant 1$. So $c-a \leqslant 7$.
But in the hundreds place $dc>d>a$, there can only be
$$
c=8, \quad a=1, \quad b=9, \quad d=3 .
$$

So the number we are looking for is
$$
\overline{a b c d}=1983 .
$$