Difference between revisions of "2017 AIME I Problems/Problem 12"

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Call a set <math>S</math> product-free if there do not exist <math>a, b, c \in S</math> (not necessarily distinct) such that <math>a b = c</math>. For example, the empty set and the set <math>\{16, 20\}</math> are product-free, whereas the sets <math>\{4, 16\}</math> and <math>\{2, 8, 16\}</math> are not product-free. Find the number of product-free subsets of the set <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}</math>.
 
Call a set <math>S</math> product-free if there do not exist <math>a, b, c \in S</math> (not necessarily distinct) such that <math>a b = c</math>. For example, the empty set and the set <math>\{16, 20\}</math> are product-free, whereas the sets <math>\{4, 16\}</math> and <math>\{2, 8, 16\}</math> are not product-free. Find the number of product-free subsets of the set <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}</math>.
  
==Solution==
+
==Solution 1(Casework)==
  
 
We shall solve this problem by doing casework on the lowest element of the subset. Note that the number <math>1</math> cannot be in the subset because <math>1*1=1</math>. Let <math>S</math> be a product-free set. If the lowest element of <math>S</math> is <math>2</math>, we consider the set <math>\{3, 6, 9\}</math>. We see that 5 of these subsets can be a subset of <math>S</math> (<math>\{3\}</math>, <math>\{6\}</math>, <math>\{9\}</math>, <math>\{6, 9\}</math>, and the empty set). Now consider the set <math>\{5, 10\}</math>. We see that 3 of these subsets can be a subset of <math>S</math> (<math>\{5\}</math>, <math>\{10\}</math>, and the empty set). Note that <math>4</math> cannot be an element of <math>S</math>, because <math>2</math> is. Now consider the set <math>\{7, 8\}</math>. All four of these subsets can be a subset of <math>S</math>. So if the smallest element of <math>S</math> is <math>2</math>, there are <math>5*3*4=60</math> possible such sets.
 
We shall solve this problem by doing casework on the lowest element of the subset. Note that the number <math>1</math> cannot be in the subset because <math>1*1=1</math>. Let <math>S</math> be a product-free set. If the lowest element of <math>S</math> is <math>2</math>, we consider the set <math>\{3, 6, 9\}</math>. We see that 5 of these subsets can be a subset of <math>S</math> (<math>\{3\}</math>, <math>\{6\}</math>, <math>\{9\}</math>, <math>\{6, 9\}</math>, and the empty set). Now consider the set <math>\{5, 10\}</math>. We see that 3 of these subsets can be a subset of <math>S</math> (<math>\{5\}</math>, <math>\{10\}</math>, and the empty set). Note that <math>4</math> cannot be an element of <math>S</math>, because <math>2</math> is. Now consider the set <math>\{7, 8\}</math>. All four of these subsets can be a subset of <math>S</math>. So if the smallest element of <math>S</math> is <math>2</math>, there are <math>5*3*4=60</math> possible such sets.
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So our answer is <math>60+64+128=\boxed{252}</math>.
 
So our answer is <math>60+64+128=\boxed{252}</math>.
  
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==Solution 2(PIE)==
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We cannot have the following pairs or triplets: <math>\{2, 4\}, \{3, 9}, \{2, 3, 6\}, \{2, 5, 10\}</math>.
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Since there are <math>512</math> subsets(<math>1</math> isn't needed) we have the following:
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<math>(512-(384-160+40-4)) \implies \boxed{252}</math>.
 
==See Also==
 
==See Also==
 
{{AIME box|year=2017|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2017|n=I|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:23, 27 June 2017

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 cannot have the following pairs or triplets: $\{2, 4\}, \{3, 9}, \{2, 3, 6\}, \{2, 5, 10\}$ (Error compiling LaTeX. ! Extra }, or forgotten $.). Since there are $512$ subsets($1$ isn't needed) we have the following: $(512-(384-160+40-4)) \implies \boxed{252}$.

See Also

2017 AIME I (ProblemsAnswer KeyResources)
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

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