Let $X$ be a subset of $\{1,2,3, \ldots, 10000\}$ with the following property: If $a, b \in X, a \neq b$, then $a \cdot b \notin X$. What is the maximal number of elements in $X$ ?

Answer: 9901.
If $X=\{100,101,102, \ldots, 9999,10000\}$, then for any two selected $a$ and $b, a \neq b$, $a \cdot b \geq 100 \cdot 101>10000$, so $a \cdot b \notin X$. So $X$ may have 9901 elements.

Suppose that $x_{1}1$ and consider the pairs

$$
\begin{gathered}
200-x_{1},\left(200-x_{1}\right) \cdot x_{1} \\
200-x_{2},\left(200-x_{2}\right) \cdot x_{2} \\
\vdots \\
200-x_{k},\left(200-x_{k}\right) \cdot x_{k}
\end{gathered}
$$

Clearly $x_{1}<x_{2}<\cdots<x_{k}<100<200-x_{k}<200-x_{k-1}<\cdots<200-x_{2}<$ $200-x_{1}<200<\left(200-x_{1}\right) \cdot x_{1}<\left(200-x_{2}\right) \cdot x_{2}<\cdots<\left(200-x_{k}\right) \cdot x_{k}$. So all numbers in these pairs are different and greater than 100. So at most one from each pair is in the set $X$. Therefore, there are at least $k$ numbers greater than 100 and $99-k$ numbers less than 100 that are not in the set $X$, together at least 99 numbers out of 10000 not being in the set $X$.