2017 AIME I Problems/Problem 12

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Solution

We shall solve this problem by doing casework on the lowest element of the subset. Note that the number $1$ cannot be in the subset because $1*1=1$. Let S be a product-free set. If the lowest element of S is $2$, we consider the set {3, 6, 9}. We see that $5$ 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 $3$ of these subsets can be a subset of S ({5}, {10}, and the empty set). Note that $4$ cannot be an element of S, because $2$ 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 $2$, there are $5*3*4=60$ possible such sets.

If the smallest element of S is $3$, the only restriction we have is that $9$ is not in S. This leaves us $2^6=64$ such sets.

If the smallest element of S is not $2$ or $3$, then S can be any subset of {4, 5, 6, 7, 8, 9, 10}, including the empty set. This gives us $2^7=128$ such subsets.

So our answer is $60+64+128=\boxed{252}$.

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