3. 22 Let $p(x)$ be the product of the digits of the decimal number $x$. Try to find all positive numbers $x$ that satisfy $p(x)=x^{2}-10 x-22$.
[Solution] Let $x$ be an $n$-digit number, then
$$p(x) \leqslant 9^{n}, \quad x^{2} \geqslant 10^{n-1} \cdot x$$

If $n>2$, then
$$\begin{aligned}
p(x)=x^{2}-10 x-22 & \geqslant x\left(10^{n-1}-10\right)-22 \\
& \geqslant 10^{n-1} \cdot 90-22=9 \cdot 10^{n}-22 \\
& >9^{n}, \text { contradiction. }
\end{aligned}$$
$$\therefore \quad n \leqslant 2 \text {. }$$

If $n=1$, then
$$\begin{array}{ll} 
& x^{2}-10 x-22=x(x-10)-221$ , then
$$\begin{aligned}
a b & >(10 a+b)(10+b)-22 \\
& >10 a b+100 a-22 \\
& >10 a b, \text { contradiction. }
\end{aligned}$$
$\therefore \quad a=1, b=(10+b) \cdot b-22$, we get $b=2$. Therefore, the positive number $x$ we are looking for is 12.