13. (52nd Moscow Mathematical Olympiad, 1989) Find all natural numbers $x$ such that the product of the digits of $x$ equals $44x - 86868$, and the sum of the digits is a perfect cube.
13. Since $44 x \geqslant 86868$, we have
$$
x \geqslant\left[\frac{86868+43}{44}\right]=1975 \text {. }
$$

Thus, $x$ must be at least a four-digit number.
On the other hand, if the number of digits $k \geqslant 5$ in $x$, then
$$
44 x-86868>4 \times 10^{k}-10^{5} \geqslant 3 \times 10^{k}>9^{k},
$$

which leads to $44 x-86868>p(x)$ (the product of $x$'s $k$ digits), a contradiction. Therefore, $x$ must be a four-digit number.
Given that the sum of the four digits of $x$, $S(x)$, satisfies $1 \leqslant S(x) \leqslant 36$, we have $S(x)=1,8$ or 27.
Clearly, $S(x)=1$ does not meet the requirement.
Since $0<p(x) \leqslant 9^{4}=6561$, we have
$$
x \leqslant\left[\frac{86868+6561}{44}\right]=2123 \text {. }
$$

The $x$ that satisfies $1975 \leqslant x \leqslant 2123$ and makes $S(x)=8$ or $27$, with $p(x) \neq 0$, are only $1989,1998,2114,2123$. Upon inspection, only $x=1989$ has the product of its digits equal to $44 x-86868$. Therefore, $x=1989$ is the unique solution to this problem.