We have assembled a large cube with $n$ edge length using $n^{3}$ unit cubes. Is there an $n$ value for which the number of small cubes cut by the space diagonals of the large cube is exactly half the number of small cubes not cut by the space diagonals?
Solution. Any space diagonal of an $n \times n \times n$ cube intersects $n$ small unit cubes. This can be easily seen if we orthogonally project the cube onto the plane of its base. The projection is an $n \times n$ square, and the projections of the space diagonals are the face diagonals.

![](https://cdn.mathpix.com/cropped/2024_05_02_1e75f39399b4d3ddc721g-1.jpg?height=219&width=225&top_left_y=270&top_left_x=945)

e.g., $n=4$

The 4 space diagonals of the cube all pass through the center of the cube.

If $n$ is even, then each diagonal passes through $n$ unit cubes, and their intersection point is a vertex of a small unit cube. Thus, there is no unit cube that is intersected by all diagonals. The number of intersected cubes is $4n$, and the number of non-intersected cubes is $8n$. According to the condition, $4n + 8n = 12n = n^3$, which gives $n^2 = 12$, which is not a solution.

![](https://cdn.mathpix.com/cropped/2024_05_02_1e75f39399b4d3ddc721g-1.jpg?height=211&width=220&top_left_y=819&top_left_x=947)

e.g., $n=5$

If $n$ is odd, then the middle small cube is intersected by all diagonals, so the number of intersected cubes is $4n - 3$; the number of non-intersected cubes is $8n - 6$, and thus $12n - 9 = n^3$. From this, $n(12 - n^2) = 9$, so $n = 3$ is a solution, meaning the $3 \times 3 \times 3$ cube satisfies the problem's requirements.