53. What is the smallest number of non-overlapping tetrahedra into which a cube can be divided?
53. It is easy to see that a cube can be divided into 5 tetrahedra. In Fig. 40, these are the tetrahedra $A A^{\prime} B^{\prime} D^{\prime}, A B^{\prime} B C, A C D D^{\prime}, B^{\prime} C^{\prime} D^{\prime} C$, and $A C D^{\prime} B^{\prime}$.

Let us now prove that it is impossible to divide the cube into fewer tetrahedra. Suppose the cube is divided into several tetrahedra. There are at least two of them whose bases lie on the face $A B C D$ of the cube. Similarly, there are at least 2 tetrahedra with bases on the face $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$.

These tetrahedra are certainly different from the first two, since a tetrahedron cannot have two parallel faces. Thus, we already have 4 tetrahedra. Their total volume is no more than $2 a^{3} / 3$, i.e., less than the volume of the cube. Therefore, it is impossible to divide the cube into 4 tetrahedra.