8. How few numbers is it possible to cross out from the sequence
$$
1,2,3, \ldots, 2023
$$
so that among those left no number is the product of any two (distinct) other numbers?
Solution. It is clear that, if we remove 43 numbers $2,3, \ldots, 44$, then, since $45^{2}=$ 2025, among those left no one is the product of any two others. This is the minimal number. To prove that consider 43 triples $(k, 89-k,(89-k) k)$, for $k=2, \ldots, 44$. They do not have numbers in common and we have to remove at least one number from every such triple.