Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2017 AIME I Problems/Problem 12

Revision as of 15:10, 27 March 2018 by Rejas (talk | contribs) (→‎Solution 2 (PIE))

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)

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 \[512 - (384 - 160 + 40 - 4) = \boxed{252}\].

See Also

2017 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_12&oldid=93554"