BRILLIAthon $2$ (Problem $2$ )

Suppose that Piercarlo's set of integers is $S$ . Any number $x$ in $S$ that is greater than $1$ must be the product of at least two primes. If $p$ is the smallest prime factor of $x$ , then $1000 \ge x \ge p^2$ , and hence $p \le 31$ . Thus $p \in P = \{2,3,5,7,11,13,17,19,23,29,31\}$ . Since all of the numbers in $S$ are coprime, there can only be one number in $S$ for each $p \in P$ . Thus $S$ can contain the number $1$ (which is not prime) and at most $|P|=11$ numbers, and hence $|S| \le 12$ .

Since Piercarlo could have chosen $\{p^2 \,|\, p \in P\} \cup \{1\} \; = \; \{1,4,9,25,49,121,169,289.361, 529, 841, 961\}$ we deduce that the greatest possible value of $|S|$ is $\boxed{12}$ .