Let $T$ be the set of all positive integer divisors of $2004^{100}$. What is the largest possible number of elements that a subset $S$ of $T$ can have if no element of $S$ is an integer multiple of any other element of $S$?
Assume throughout that \(a, b, c\) are nonnegative integers. Since the prime factorization of 2004 is \(2004=2^{2} \cdot 3 \cdot 167\),

\[
T=\left\{2^{a} 3^{b} 167^{c} \mid 0 \leq a \leq 200, 0 \leq b, c \leq 100\right\}
\]

Let

\[
S=\left\{2^{200-b-c} 3^{b} 167^{c} \mid 0 \leq b, c \leq 100\right\}
\]

For any \(0 \leq b, c \leq 100\), we have \(0 \leq 200-b-c \leq 200\), so \(S\) is a subset of \(T\). Since there are 101 possible values for \(b\) and 101 possible values for \(c\), \(S\) contains \(101^{2}\) elements. We will show that no element of \(S\) is a multiple of another and that no larger subset of \(T\) satisfies this condition.

Suppose \(2^{200-b-c} 3^{b} 167^{c}\) is an integer multiple of \(2^{200-j-k} 3^{j} 167^{k}\). Then

\[
200-b-c \geq 200-j-k, \quad b \geq j, \quad c \geq k
\]

But this first inequality implies \(b+c \leq j+k\), which together with \(b \geq j, c \geq k\) gives \(b=j\) and \(c=k\). Hence no element of \(S\) is an integer multiple of another element of \(S\).

Let \(U\) be a subset of \(T\) with more than \(101^{2}\) elements. Since there are only \(101^{2}\) distinct pairs \((b, c)\) with \(0 \leq b, c \leq 100\), then (by the pigeonhole principle) \(U\) must contain two elements \(u_{1}=2^{a_{1}} 3^{b_{1}} 167^{c_{1}}\) and \(u_{2}=2^{a_{2}} 3^{b_{2}} 167^{c_{2}}\), with \(b_{1}=b_{2}\) and \(c_{1}=c_{2}\), but \(a_{1} \neq a_{2}\). If \(a_{1}>a_{2}\), then \(u_{1}\) is a multiple of \(u_{2}\) and if \(a_{1}<a_{2}\), then \(u_{2}\) is a multiple of \(u_{1}\). Hence \(U\) does not satisfy the desired condition.

Therefore the largest possible number of elements that such a subset of \(T\) can have is \(101^{2}=10201\).