2017 AIME I Problems/Problem 12

Problem 12

Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ .

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 $60+64+128=\boxed{252}$ .

2017 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2017 AIME I Problems/Problem 12

Problem 12

Solution

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