# 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 $\{1, 2, 3, 4,..., 7, 8, 9, 10\}$.

## 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)

We will consider the $2^9 = 512$ subsets that do not contain 1. A subset is product-free if and only if it does not contain one of the groups $\{2, 4\}, \{3, 9\}, \{2, 3, 6\},$ or $\{2, 5, 10\}$. There are $2^7$ subsets that contain 2 and 4 since each of the seven elements other than 2 and 4 can either be in the subset or not. Similarly, there are $2^7$ subsets that contain 3 and 9, $2^6$ subsets that contain 2, 3, and 6, and $2^6$ subsets that contain 2, 5, and 10. The number of sets that contain one of the groups is: $$2^7 + 2^7 + 2^6 + 2^6 = 384$$ For sets that contain two of the groups, we have: $$2^5 + 2^5 + 2^5 + 2^5 + 2^4 + 2^4 = 160$$ For sets that contain three of the groups, we have: $$2^4 + 2^3 + 2^3 + 2^3 = 40$$ For sets that contain all of the groups, we have: $$2^2 = 4$$ By the principle of inclusion and exclusion, the number of product-free subsets is $$512 - (384 - 160 + 40 - 4) = \boxed{252}$$.

## Solution 3

Let $X$ be a product-free subset, and note that 1 is not in $x$. We consider four cases:

1.) both 2 and 3 are not in $X$. Then there are $2^7=128$ possible subsets for this case.

2.) 2 is in $X$, but 3 is not. Then 4 in not in $X$, so there are $2^6=64$ subsets; however, there is a $\frac{1}{4}$ chance that 5 and 10 are both in $X$, so there are $64\cdot \frac{3}{4}=48$ subsets for this case.

3.) 2 is not in $X$, but 3 is. Then, 9 is not in $X$, so there are $2^6=64$ subsets for this case.

4.) 2 and 3 are both in $X$. Then, 4, 6, and 9 are not in $X$, so there are $2^4=16$ total subsets; however, there is a $\frac{1}{4}$ chance that 5 and 10 are both in $X$, so there are $16\cdot \frac{3}{4}=12$ subsets for this case.

Hence our answer is $128+48+64+12=\boxed{252}$. -Stormersyle

## Solution 4

First, ignore 7 as it does not affect any of the other numbers; we can multiply by 2 later (its either in or out).

Next, we do casework based on whether 2, 3, or 5 are in the set ( $2^3 = 8$ quick cases). For each of the cases, the numbers will be of two types ---

1. They cannot be in the set because they form a product

2. Whether we add the number to the set or not, it does not affect the rest of the numbers. These numbers simply add a factor of 2 to the case.

For example, in the case where all three are present, we can see that $4, 6, 9, 10$ are of type 1. However, $8$ is of type 2. Therefore this case contributes 2.

Summing over the 8 cases we get $2 + 4 + 8 + 16 + 16 + 16 + 32 + 32 = 126.$ Multiplying by $2$ because of the $7$ gives $252.$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 