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 1(Casework)

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

Solution 2(PIE) (Should be explained in more detail)

We cannot have the following pairs or triplets: $\{2, 4\}, \{3, 9\}, \{2, 3, 6\}, \{2, 5, 10\}$ . Since there are $2^9$ = $512$ subsets( $1$ isn't needed) we have the following: The total number of sets with at least one of the groups (with repeats) = $2^7 + 2^7 + 2^6 + 2^6$ = $384$ The total number of sets with at least two of the groups (with repeats) = 1st & 2nd pair $+$ 1st & 3rd pair $+$ 1st & 4th pair $+$ 2nd & 3rd pair $+$ 2nd & 4th pair $+$ 3rd & 4th pair = $2^5 + 2^5 + 2^5 + 2^5 + 2^4 + 2^4$ = $160$ The total number of sets with at least three of the groups (with repeats) = 1st, 2nd, & 3rd groups $+$ 1st, 2nd & 4th groups $+$ 1st, 3rd, & 4th groups $+$ 2nd, 3rd, & 4th groups = $2^4 + 2^3 + 2^3 + 2^3$ = $40$ The total number of sets with all of the groups. = $2^2$ = $4$ $(512-(384-160+40-4)) \implies \boxed{252}$ .

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

Contents

Problem 12

Solution 1(Casework)

Solution 2(PIE) (Should be explained in more detail)

See Also