10.64 For a natural number $A$, define the product of its (decimal) digits as $p(A)$. Find all $A$ that satisfy $A=1.5 p(A)$.
(China National Training Team Problem, 1990)
［Solution］Let $A$ be an $n$-digit number $(n \geqslant 2)$, with its first digit being $a$, then we have
$$
A \geqslant a \cdot 10^{n-1} \text {. }
$$

If all other digits of $A$ are 9, then $9 \mid p(A)$, and from
$$
A=1.5 p(A)
$$

we get
$$
9 \mid A \text {. }
$$

Thus,
$$
9 \mid a \text {. }
$$

At this point,
$$
1.5 p(A)=\frac{3}{2} \cdot 9^{n}
$$

is not an integer, so not all digits after the first digit of $A$ are 9, hence
$$
p(A) \leqslant 8 a \cdot 9^{n-2} \text {. }
$$

From (1) and (2), we get

Thus,
$$
\begin{array}{l}
1.5 \cdot 8 a \cdot 9^{n-2} \geqslant a \cdot 10^{n-1} \\
12 \cdot 9^{n-2} \geqslant 10^{n-1}, \quad n \leqslant 3 .
\end{array}
$$

If $n=2$, let $A=10 a+b$, then

Solving, we get
$$
\begin{array}{l}
20 a+2 b=3 a b, \\
a=4, \quad b=8 .
\end{array}
$$

Thus,
$$
A=48 \text {. }
$$

If $n=3$, let $A=100 a+10 b+c$, then
$$
200 a+20 b+2 c=3 a b c .
$$

Therefore,
$$
b c>\frac{200}{3}>66 \text {. }
$$
$b$ and $c$ can only be $(9,9)$ or $(9,8)$. In both cases, $a$ is not an integer.
So the only solution to this problem is: $A=48$.