Example 43 Let $T$ be the set of all positive divisors of $60^{100}$. $S$ is a subset of $T$ in which no number is a multiple of another. Find the maximum value of $|S|$.


Assume that $a, b, c$ are all non-negative integers. Since $60=2^{2} \cdot 3 \cdot 5$,
$$
T=\left\{2^{a} 3^{b} 5^{c} \mid 0 \leqslant a \leqslant 200,0 \leqslant b, c \leqslant 100\right\},
$$

Let $\quad S=\left\{2^{200-b-c} 3^{b} 5^{c} \mid 0 \leqslant b, c \leqslant 100\right\}$,
For any $0 \leqslant b, c \leqslant 100,0 \leqslant 200-b-c \leqslant 200$, so $S$ is a subset of $T$ and contains $101^{2}$ elements. Below we prove that no element in $S$ is a multiple of another, and any other subset with more than $101^{2}$ elements does not satisfy this condition.
Assume $2^{200 \cdots \cdots} 3^{b} 5^{c}$ is a multiple of $2^{200 \cdots i-j} 3^{i} 5^{j}$, then
$$
200-b-c \geqslant 200-i-j, b \geqslant i, c \geqslant j
$$

The first inequality indicates $b+c \leqslant i+j$, combined with the last two inequalities we get $b=i, c=j$. Therefore, no element in $S$ is a multiple of another.

Let $U$ be a subset of $T$ with more than $101^{2}$ elements. Since there are only $101^{2}$ pairs of distinct $(b, c)$, by the pigeonhole principle, $U$ must contain two elements $u_{1}=2^{a_{1}} 3^{b_{1}} 5^{c_{1}}, u_{2}=2^{a_{2}} 3^{b_{2}} 5^{c_{2}}$, where $b_{1}=b_{2}, c_{1}=c_{2}$, and $a_{1} \neq a_{2}$. If $a_{1}>a_{2}$, then $u_{1}$ is a multiple of $u_{2}$; if $a_{1}<$ $a_{2}$, then $u_{2}$ is a multiple of $u_{1}$. Therefore, $U$ does not satisfy the given condition.

Thus, the subset of $T$ can contain at most $101^{2}=10201$ elements. In solving specific problems, how to apply this to solve the problem. In specific problems, how to "create" pigeonholes, through the above examples, it is hoped that readers will gain some insight.