310 Let $n$ be a given positive integer. $D_{n}$ is the set of all divisors of $2^{n} 3^{n} 5^{n}$. $S \subseteq D_{n}$. If no number in $S$ can divide any other number in $S$, find the maximum value of $|S|$.
Solve $|S|_{\max }=\left[\frac{3(n \pm 1)^{2}}{4} \pm 1\right]$.
Obviously, each number in $D_{n}$ has the form $2^{\alpha} 3 \beta-\gamma \gamma$, where $0 \leqslant \alpha, \beta$, $\gamma \leqslant n$. Hereafter, use the array $(\alpha, \beta, \gamma)$ to represent the number $2 \alpha 3 \beta 5 \gamma$.
Consider the following set:
$$
\begin{aligned}
A_{i, j}= & \{(i, j, \alpha), 0 \leqslant \alpha \leqslant n-j\} \cup \\
& \{(i, \beta, n-j), j \leqslant \beta \leqslant n\} .
\end{aligned}
$$
$$
\begin{array}{c}
i=0,1,2, \cdots, n, j=0,1,2, \cdots,\left[\frac{n}{2}\right] ; \\
B_{i, j}=\{(k, i, j), k=0,1,2, \cdots, n\} \\
i=\left[\frac{n^{-}}{2}+1, \cdots, n, j=0,1,2, \cdots, n-\left[\frac{n}{2}\right]-1 .\right.
\end{array}
$$

In total, there are
$$
\begin{aligned}
& \left(\left[\frac{n}{2}-1\right) \cdot(n+1)+\left(n-\left[\frac{n}{2}\right]\right)^{2}\right. \\
= & \left(\frac{n}{2}+1\right)(n+1)+\frac{n^{2}}{4} \\
= & \frac{3(n+1)^{2}+1}{4}(n \text { is even). } \\
& \left(\left[\frac{n^{-}}{2}+1\right)(n+1)+\left(n-\left[\frac{n}{2}\right]\right)^{2}\right. \\
= & \left(\frac{n-1}{2}+1\right)(n+1)+\frac{(n+1)^{2}}{4} \\
- & \frac{3(n+1)^{2}}{4}(n \text { is odd). }
\end{aligned}
$$

Thus,
$$
\begin{aligned}
& \left(\left[\frac{n^{-}}{2}+1\right)(n+1)+\left(n-\left[\frac{n}{2}\right]\right)^{2}\right. \\
= & \left(\frac{n-1}{2}+1\right)(n+1)+\frac{(n+1)^{2}}{4} \\
- & \frac{3(n+1)^{2}}{4}(n \text { is odd) } .
\end{aligned}
$$

In summary, there are $\left[\frac{3(n+1)^{2}+1}{4}\right]$ sets.
Since each set contains at most one number belonging to $\mathrm{f}$ (because any two numbers in each set have a multiple relationship), we can conclude that the sets are mutually exclusive and cover all the numbers. Thus, $\cup A_{i,}, \cup B_{i .,}=I_{n}$. Therefore, we have $|S| \leqslant\left[\frac{3(n+1)^{2}+1}{4}\right]$.

On the other hand, consider the number of arrays $(\alpha, \beta, \gamma)$ satisfying $\alpha+\beta+\gamma=n+t\left(\right.$ where $\left.t=\left[\frac{n}{2}\right]\right)$.
Given the condition $0 \leqslant \alpha, \beta, \gamma \leqslant n$, we get
$$
\begin{aligned}
& (n+t+1)+(n+t)+\cdots+1-3 t+(t-1)+\cdots+1] \\
= & \frac{1}{2}(n+t+2)(n+t+1)-\frac{3}{2} t(t+1) \\
= & \frac{1}{2}(n+1)(n+2)+\frac{1}{4} n^{2}-\left(t-\frac{n}{2}\right)^{2} \\
= & \frac{1}{4}\left[3(n+1)^{2}+1\right]-\left(t-\frac{n}{2}\right)^{2} \\
= & {\left[\frac{3(n+1)^{2}+1}{4}\right] \cdot\left(\text { where } t=\left[\frac{n}{2}\right]\right) }
\end{aligned}
$$
(When $\alpha, \beta, \gamma$ is greater than $n$, there are $3(t+(t-1)+\cdots+1)$ solutions, so they are excluded)

Let $S=\left\{203 \beta-\left.\gamma\right|_{\alpha}+\beta+\gamma=n+\left[\frac{n}{2}\right], 0 \leqslant \alpha, \beta \cdot \gamma \leqslant n\right\}$. From this, we get $|S|=\left[\frac{3(n+1)}{4}+1\right]$. H. $S$ contains no two numbers with a multiple relationship. Suppose $\left.2^{\alpha_{1}} 3^{\beta_{1}}-\right)^{\gamma} \mid 2^{\alpha_{1}}=3^{\gamma^{\prime}} 55^{\gamma}$. Then $\alpha_{1} \leqslant \alpha_{2} \cdot \beta_{1} \leqslant \beta_{2}, \gamma_{1} \leqslant \gamma_{2}$, $\alpha_{1}+\beta_{1}+\gamma_{1}=\alpha_{2}+\beta_{2}+\gamma_{2}$. Therefore, $\alpha_{1}=\alpha_{2}, \beta_{1}=\beta_{2}, \gamma_{1}=\gamma_{2}$, leading to a contradiction!
In summary, we conclude $|S|_{\max }=\left[\frac{3(n+1)^{2}+1}{1}\right]$.