Which are those natural numbers that are eleven times as large as the product of their digits?
Let the product of the digits of the number $A$ written in the decimal system be denoted by $P(A)$. If the first digit of the $n$-digit number $A$ is $a$ $(n \geq 2)$, then

$$
A \geq a \cdot 10^{n-1}
$$

If the other digits of $A$ are 9s, then $9 \mid P(A)$ and thus if

$$
A=1.5 \cdot P(A)
$$

then $9 \mid A$, which implies $9 \mid a$, so $a=9$ and $A=10^{n}-1$, but in this case $1.5 P(A)$ is not an integer. Therefore, among the other digits of $A$, there is one that is less than 9, so $P(A) \leq 8 \cdot 9^{n-2} \cdot a$. Comparing this with (1) and (2),

$$
\begin{gathered}
1.5 \cdot 8 \cdot 9^{n-2} \geq 10^{n-1} \\
\frac{6}{5} \geq\left(\frac{10}{9}\right)^{n-2}
\end{gathered}
$$

Since the right side increases monotonically with $n$ and $\left(\frac{10}{9}\right)^{2}>\frac{6}{5}$, from (3) we get $n-2 \leq 1$, i.e., $n \leq 3$.

If $n=1$, then $A=0$. If $n=2$, then

$$
A=10 a+b=1.5 a b
$$

From this, $3 a b-20 a-2 b=0$, so

$$
b=\frac{20 a}{3 a-2}=6+\frac{2 a+12}{3 a-2}
$$

which means $b \geq 7$. If $b=7$, then $a>9$, if $b=8$, then $a=4$ is a solution, and if $b=9$, then $a$ is not an integer.

If $n=3$, then $A=100 a+10 b+c=1.5 a b c$, so

$$
200 a+20 b+2 c=3 a b c
$$

which means

$$
200 a66
$$

Thus, either $b=c=9$, or $\{b, c\}=\{9,8\}$.

It is easy to see that in neither case do we get a solution, which means that the problem has two solutions: $A=48$ and $A=0$.