The 72nd question: Given a positive integer $\mathrm{n}$, a regular hexagon with side length $\mathrm{n}$ is divided into a series of equilateral triangles with side length 1, parallel to the sides of the hexagon. Find the number of regular hexagons whose vertices are the vertices of these equilateral triangles.
The 72nd question:
Answer: We call a regular hexagon a standard hexagon if its vertices are the vertices of small equilateral triangles, and its six sides are parallel to the six sides of the large hexagon. For any regular hexagon, we can find a standard hexagon such that the six vertices of the regular hexagon lie on the six sides of the standard hexagon. Conversely, for any standard hexagon with a long side length of \( \mathrm{m} \), there are exactly \( \mathrm{m} \) regular hexagons inscribed in this standard hexagon.

Next, we calculate the number of standard hexagons with a side length of \( \mathrm{m} \). Since the center of such a standard hexagon is no more than \( \mathrm{n}-\mathrm{m} \) units away from the center of the large hexagon, we only need to count the lattice points within a regular hexagon centered at the center of the large hexagon with a side length of \( \mathrm{n}-\mathrm{m} \). Therefore, the number of standard hexagons with a side length of \( \mathrm{m} \) is:
$$
6\left(\sum_{\mathrm{i}=1}^{\mathrm{n}-\mathrm{m}} \mathrm{i}\right)+1=3(\mathrm{n}-\mathrm{m})(\mathrm{n}-\mathrm{m}+1)+1
$$

Thus, the number of regular hexagons that meet the conditions is:
$$
\begin{array}{l}
\sum_{i=1}^{n}[3(n-m)(n-m+1)+1] m \\
=\sum_{i=1}^{n}\left[\left(3 n^{2}+3 n+1\right)-3(2 n+1) m+3 m^{2}\right] m \\
=\frac{n^{2}(n+1)^{2}}{4}
\end{array}
$$

In conclusion, the number of regular hexagons that meet the conditions is \( \frac{\mathrm{n}^{2}(n+1)^{2}}{4} \).