What is the least number of unit cubes needed to build a larger cube such that more than half of the small cubes are inside?
To have a small cube inside a large cube, at least $3 \times 3 \times 3=27$ cubes must be assembled. This is the smallest possibility. Let's examine how the number of inner cubes changes as we increase the sides of the large cube. For this, let's create the following table.

Figure 2

| Total Cubes |
| :--- |
| $3 \times 3 \times 3=27$ |
| $4 \times 4 \times 4=64$ |
| $5 \times 5 \times 5=125$ |
| $6 \times 6 \times 6=216$ |
| $\vdots$ |

From Figure 2, it can be seen that for an $n \times n \times n$ cube, the number of inner cubes is $(n-2) \times (n-2) \times (n-2)$, since in each layer, the outermost rows must be left out, and these will cover the inner cubes. Therefore, the number of inner cubes is $(n-2)^{3}$. The question is, when will $(n-2)^{3} > \frac{1}{2} n^{3}$.

Rearranging the inequality:

$$
\begin{aligned}
& n-2 > n \sqrt[3]{\frac{1}{2}} \\
& n > \frac{2}{1-\sqrt[3]{\frac{1}{2}}} = \frac{2 \sqrt[3]{2}}{\sqrt[3]{2}-1} \approx 9.69
\end{aligned}
$$

Since $n$ is an integer, $n$ must be at least 10, meaning that in the case of a large cube assembled from $10 \times 10 \times 10=1000$ small cubes, the number of inner cubes is $8 \times 8 \times 8=512$, which is indeed greater than half of 1000.

![](https://cdn.mathpix.com/cropped/2024_05_02_4cb42afd15c7406acf15g-1.jpg?height=648&width=639&top_left_y=1324&top_left_x=732)

![](https://cdn.mathpix.com/cropped/2024_05_02_4cb42afd15c7406acf15g-2.jpg?height=1499&width=1499&top_left_y=129&top_left_x=259)