10.63 Let $p(x)$ be the product of the digits of the decimal integer $x$. Find all positive integers $x$ such that $p(x)=x^{2}-10 x-22$.
(10th International Mathematical Olympiad, 1968)
[Solution 1] Let the decimal natural number $x$ consist of $n$ digits. Then $p(x)$ and $x$ should satisfy
$$
p(x) \leqslant 9^{n}, \quad x \geqslant 10^{n-1}.
$$
(1) When $n=1$, then $p(x)=x$, thus
$$
\begin{array}{l}
x^{2}-10 x-22=x, \\
x^{2}-11 x-22=0.
\end{array}
$$

This equation has no integer solutions.
(2) When $n=2$, then we have

Thus,
$$
\begin{array}{c}
x^{2}-10 x-22=p(x) \leqslant 81, \\
x^{2}-10 x+25=p(x)+47 \leqslant 128, \\
(x-5)^{2}=p(x)+47 \leqslant 128 \\
|x-5| \leqslant \sqrt{128}2$ when $n>2$, from $x \geqslant 10^{n-1}$ we get
$$
\begin{array}{c}
02$ we know
$$
10^{n} \geqslant 1000, \quad 10^{n-2}-2 \geqslant 8.
$$

Therefore,
$$
\begin{array}{l}
10^{n}\left(10^{n-2}-2\right) \geqslant 8000, \\
p(x) \geqslant 10^{2 n-2}-10^{n}-22 \\
=10^{n}\left(10^{n-2}-2\right)+10^{n}-22 \\
\geqslant 8000+10^{n}-22 \\
>10^{n},
\end{array}
$$

Thus,
$$
\begin{aligned}
p(x) & \geqslant 10^{2 n-2}-10^{n}-22 \\
& =10^{n}\left(10^{n-2}-2\right)+10^{n}-22 \\
& \geqslant 8000+10^{n}-22 \\
& >10^{n}
\end{aligned}
$$

which contradicts $p(x) \leqslant 9^{n}$.
So there is no solution when $n>2$.
From (1), (2), and (3), we know that the only positive integer solution is $x=12$.
[Solution 2] Let the decimal natural number $x$ consist of $n$ digits. Then $p(x) \leqslant 9^{n}$.
(1) When $n=1$, the same as Solution 1, there is no solution.
(2) When $n \geqslant 2$, then
$$
x \geqslant 10^{n-1} a_{n}>9^{n-1} a_{n} \geqslant p(x),
$$

where $a_{n}$ is the $n$-th digit of $x$.
So we have
$$
\begin{array}{c}
p(x)=x^{2}-10 x-220$ we get
$$
0<x<\frac{11+\sqrt{209}}{2}<13.
$$

Since $x$ is a natural number, and $x \geqslant 10$, then
$$
x=10,11,12.
$$

If $x=10$, then $p(x)=0$,
$$
x^{2}-10 x-22=-22 \neq p(x).
$$

If $x=11$, then $p(x)=1$,
$$
x^{2}-10 x-22=-11 \neq p(x).
$$

If $x=12$, then $p(x)=2$,
$$
x^{2}-10 x-22=2=p(x).
$$

Therefore, the natural number that meets the requirement of this problem is only $x=12$.