$1 \cdot 25$ Find all natural numbers $x$ that satisfy the following conditions: the product of the digits of $x$ equals $44x-86868$, and the sum of the digits is a perfect cube.
[Solution] Since $44 x \geqslant 86868$, hence $x \geqslant\left[\frac{86868+43}{44}\right]=1975$.
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
$$44 x-86868>4 \times 10^{k}-10^{5} \geqslant 3 \times 10^{k}>9^{k}$$

Thus, we get $44 x-86868>p(x)$ (the product of $x$'s $k$ digits), which is a contradiction, so $x$ must be a four-digit number.

From the given, the sum of the four digits of $x$, $S(x)$, satisfies $1 \leqslant S(x) \leqslant 36$. Therefore, $S(x)=$ 1, 8, or 27.

Clearly, $S(x)=1$ does not meet the requirements.
Since $0<p(x) \leqslant 9^{4}=6561$,
we have $x \leqslant\left[\frac{86868+6561}{44}\right]=2123$.
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. After inspection, only $x=1989$ has the product of its digits equal to $44 x-86868$, so $x=1989$ is the only solution to this problem.