1.104 Divide each edge of a regular tetrahedron into 5 equal parts, and through each division point, draw two planes, each parallel to the two faces of the tetrahedron that do not pass through that point. How many parts do these planes divide the tetrahedron into?
[Solution] (1) First, consider the case when each edge of the tetrahedron is divided into 2 equal parts and planes are drawn as required. It is easy to see that this results in 4 smaller tetrahedra and 1 octahedron, a total of 5 parts.
(2) Next, consider the case when each edge of the tetrahedron is divided into 3 equal parts and planes are drawn as required. In this case, removing the bottom layer leaves a regular tetrahedron with each edge divided into 2 equal parts. We call this a "trimmed tetrahedron." After doing this for each side, we get 4 trimmed tetrahedra. From (1), we know that each trimmed tetrahedron is divided into 5 parts, totaling 20 parts.

On the other hand, the intersection of any two trimmed tetrahedra is exactly one small tetrahedron, so among the 20 parts, 6 small tetrahedra are counted twice. Thus, the non-repeated parts total 14. Additionally, the original tetrahedron, after removing the 4 trimmed tetrahedra, leaves 1 small tetrahedron, which has its 4 vertices at the centers of the original tetrahedron's 4 faces. Therefore, the tetrahedron is divided into 15 parts.
(3) Then, consider the case when each edge of the tetrahedron is divided into 4 equal parts and planes are drawn as required. The tetrahedron obtained by removing the bottom layer is still called a trimmed tetrahedron. From (2), we know that each trimmed tetrahedron is divided into 15 parts. At the same time, the intersection of any two trimmed tetrahedra is a 2-layer tetrahedron, and the intersection of any three trimmed tetrahedra is a small tetrahedron. Therefore, by the principle of inclusion-exclusion, we have
$$
p_{4}=15 \times 4-5 \times 6+4=60-30+4=34,
$$

which means the tetrahedron is divided into 34 parts.
(4) Finally, consider the case when each edge of the tetrahedron is divided into 5 equal parts and planes are drawn as required. In this case, the edges of each trimmed tetrahedron are divided into 4 equal parts, the intersection of any two trimmed tetrahedra is a 3-layer tetrahedron, the intersection of any three trimmed tetrahedra is a 2-layer tetrahedron, and the intersection of any four trimmed tetrahedra is a small tetrahedron. Thus, by the principle of inclusion-exclusion, we have
$$
p_{5}=34 \times 4-15 \times 6+5 \times 4-1=65,
$$

which means the tetrahedron is divided into 65 parts.