Find all natural numbers $x$, satisfying the conditions: the product of the digits of the number $x$ is equal to $44x - 86868$, and the sum of the digits is a cube of a natural number.
Let $\overline{a_{n}^{\ldots}-\ldots a_{0}^{-}}$ be the decimal representation of the number $a$. Then $a \geq a_{n} \cdot 10^{n} \geq a_{n} a_{n-1} \cdot 9^{n} \geq a_{n} a_{n-1} \ldots a_{0}$. From this it follows that $x \geq$ $44 x$ - 86868, from which $x \leq 2020$. On the other hand, the product of the digits of any number is non-negative, so $44 x \geq 86868$, from which $x \geq 1975$. For numbers from 2000 to 2020, the product of the digits is zero, which cannot be equal to $44 x$ - 86868 (since 86868 is not divisible by 44). Therefore, $1975 \leq x \leq 1999$, which means the sum of the digits of the number $x$ is not less than $1+9=10$ and not more than $1+9+9+9=28$. In this range, there is only one perfect cube, so the sum of the digits of the number $x$ is 27, which means either $x=1989$ or $x=1998$. In both cases, the product of the digits is $1 \cdot 9 \cdot 9 \cdot 8=648$. Therefore, $44 x=86868+648=87516$, from which $x=1989$.