4. The product of the number 21 with some four-digit number $x$ is the cube of some integer $y$. Find the number $x$.

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Solution. From the condition of the problem, we have $21 x=y^{3}$, from which it follows that 21 is a divisor of $y$, i.e., $y=21 z$. Substituting into $21 x=y^{3}$, we get $x=441 z^{3}$. The number $x$ is a four-digit number, so $z$ can be either 1 or 2, because for $z \geq 3$ we have $x \geq 441 \cdot 27=11907$. For $z=1$ we have $x=441$, and for $z=2$ we have $x=3528$. Therefore, $x=3528$, and in this case $21 x=168^{3}$.