Let's divide a solid cube with an edge of $n \mathrm{~cm}$ into smaller cubes with an edge of $1 \mathrm{~cm}$ ( $n$ is a natural number). For which $n$ can a cube with an edge of $2 n \mathrm{~cm}$ be assembled from the resulting small cubes, if we allow there to be a void inside the cube?
After the dissection of an $n \mathrm{~cm}$ edge, it breaks down into $n^{3}$ number of $1 \mathrm{~cm}$ edge cubes. The $2 n \mathrm{~cm}$ edge cube can be assembled according to the conditions if a small cube fits on each $1 \mathrm{~cm}$ edge square that forms its surface. This is true when $n^{3} \geq(2 n)^{3}-(2 n-2)^{3}$, that is, $(2 n-2)^{3} \geq 7 n^{3}$. Since the function $x \mapsto \sqrt[3]{x}$ is bijective, the previous inequality is true precisely when $2 n-2 \geq n \sqrt[3]{7}$, from which $n \geq \frac{2}{2-\sqrt[3]{7}} \approx 22.97$. Therefore, the $2 n \mathrm{~cm}$ edge cube can be assembled in the desired manner if $n$ is at least 23.

Bajusz Csaba (Szeged, Radnóti M. Gymn., 10th grade)