# 2017 AIME I Problems/Problem 12

## Solution

We shall solve this problem by doing casework on the lowest element of the subset. Note that the number cannot be in the subset because . Let S be a product-free set. If the lowest element of S is , we consider the set {3, 6, 9}. We see that of these subsets can be a subset of S ({3}, {6}, {9}, {6, 9}, and the empty set). Now consider the set {5, 10}. We see that of these subsets can be a subset of S ({5}, {10}, and the empty set). Note that cannot be an element of S, because is. Now consider the set {7, 8}. All four of these subsets can be a subset of S. So if the smallest element of S is , there are possible such sets.

If the smallest element of S is , the only restriction we have is that is not in S. This leaves us such sets.

If the smallest element of S is not or , then S can be any subset of {4, 5, 6, 7, 8, 9, 10}, including the empty set. This gives us such subsets.

So our answer is .