7.49 Divide a cube into non-overlapping tetrahedra. What is the minimum number of tetrahedra needed?
[Solution] Each face of a tetrahedron is a triangle, while the 6 faces of a cube are all squares, and each square can be divided into at least two triangles, resulting in a total of at least 12 triangles. Among the 4 faces of the divided tetrahedra, at most 3 faces can be on the surface of the cube, meaning each tetrahedron can use at most 3 of the at least 12 triangles on the cube, so there must be at least 4 such tetrahedra.

If the edge length of the cube is denoted as $a$, then the area of each triangle on the surface of the cube does not exceed $\frac{1}{2} a^{2}$, so the volume of each tetrahedron does not exceed $\frac{1}{6} a^{3}$. Therefore, 4 tetrahedra cannot fill the cube. It follows that the cube must be divided into at least 5 tetrahedra.

As shown in the figure, this is an example of dividing the cube into 5 tetrahedra. Thus, the minimum number of tetrahedra into which a cube can be divided is 5.