4. Let's call a rectangular parallelepiped typical if all its dimensions (length, width, and height) are different. What is the smallest number of typical parallelepipeds into which a cube can be cut? Don't forget to prove that this is indeed the smallest number.
Solution. A cube can be cut into four typical parallelepipeds. For example, a cube $5 \times 5 \times 5$ can be cut into parallelepipeds $5 \times 3 \times 1, 5 \times 3 \times 4, 5 \times 2 \times 1, 5 \times 2 \times 4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bfb93b3716727f813bd7g-2.jpg?height=401&width=401&top_left_y=808&top_left_x=819)

It is impossible to cut the cube into a smaller number of typical parallelepipeds. Indeed, a cube has 8 vertices; if it is cut into three or fewer parallelepipeds, then at least one of them contains at least three vertices of the cube.

If all three vertices are located on the same face of the cube (for example, on the top face), then the parallelepiped contains the entire top face of the cube; thus, it has two identical dimensions. If two vertices are located on the top face of the cube and one on the bottom, then the parallelepiped contains at least one edge of the top face, and its height is equal to the height of the cube. Again, we have two identical dimensions.

Answer: into 4 typical parallelepipeds.