7. There are 120 equally distributed spheres inside a regular tetrahedron $A-B C D$. How many spheres are placed at the bottom of this regular tetrahedron?
Analysis (1) The base of the regular tetrahedron $A-BCD$ is the equilateral $\triangle BCD$. Assuming the number of spheres closest to the edge $BC$ is $n$, then the number of rows of spheres tangent to the base $\triangle BCD$ is also $n$, with the number of spheres in each row being $n, n-1, \cdots, 3,2,1$. Thus, the total number of spheres tangent to the base is
$$
1+2+\cdots+n=\frac{n(n+1)}{2} \uparrow .
$$
The number of spheres in each layer is
$$
\begin{array}{l}
1+2+\cdots+(n-1)=\frac{(n-1) n}{2} . \\
1+2+\cdots+(n-2)=\frac{(n-2)(n-1)}{2} . \\
\cdots \cdots \\
1+2=\frac{2 \times 3}{2} . \\
1=\frac{1 \times 2}{2} . \\
\therefore \frac{n(n+1)}{2}+\frac{(n-1) n}{2}+\cdots+\frac{1 \times 2}{2}=120 .
\end{array}
$$

That is, $\frac{1}{6} n(n+1)(n+2)=120$.
$$
\begin{array}{l}
(n-8)\left(n^{2}+11 n+90\right)=0 . \\
\therefore n=8 .
\end{array}
$$

Therefore, a total of 8 small spheres are placed inside the regular tetrahedron, and the number of spheres at the bottom is $\frac{8 \times 9}{2}=36$.
（3）The problem of densely packing small spheres inside a regular tetrahedron: From top to bottom, the 1st layer has 1 sphere, the 2nd layer has 3 spheres, the 3rd layer has 6 spheres, $\cdots$, the $n$-th layer has $\frac{1}{2} n(n+1)$ spheres. The number of spheres placed in each layer is equal to half the product of the layer number and the next layer number.