What is the smallest number of non-overlapping tetrahedra into which a cube can be divided?

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Answer: 5. If a tetrahedron $A^{\prime} B C^{\prime} D$ is cut out from the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, the remaining part of the cube splits into 4 tetrahedra, i.e., the cube can be cut into 5 tetrahedra. We will prove that it is impossible to cut the cube into fewer than 5 tetrahedra. The face $A B C D$ cannot be a face of the tetrahedra into which the cube is divided, because at least two tetrahedra adjoin it. Consider all tetrahedra adjacent to the face $A B C D$. Their heights dropped onto this face do not exceed $a$, where $a$ is the edge of the cube, and the sum of the areas of their faces lying on $A B C D$ is equal to $a^{2}$. Therefore, the sum of their volumes does not exceed $a^{3} / 3$. Since the faces of one tetrahedron cannot lie on opposite faces of the cube, at least 4 tetrahedra adjoin the faces $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, and the sum of their volumes does not exceed $2 a^{3} / 3 < a^{3}$. Therefore, there is still one more tetrahedron in the partition.