15.22. What is the smallest number of tetrahedra into which a cube can be cut?
15.22. 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 the cube cannot be cut into fewer than 5 tetrahedra. The face \(A B C D\) cannot be a face of the tetrahedra into which the cube is divided, so 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.