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

m (→‎Solution 2(PIE) (Should be explained in more detail))
(→‎Solution 2(PIE) (Should be explained in more detail))
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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) (Should be explained in more detail)==
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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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We will consider the <math>2^9 = 512</math> subsets that do not contain <math>1</math>. A subset is product-free if and only if it does not contain one of the groups <math>\{2, 4\}, \{3, 9\}, \{2, 3, 6\},</math> or <math>\{2, 5, 10\}</math>. There are <math>2^7</math> 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 <math>2^7</math> subsets that contain 3 and 9, <math>2^6</math> subsets that contain 2, 3, and 6, and <math>2^6</math> subsets that contain 2, 5, and 10. The number of sets that contain one of the groups is:
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Since there are <math>2^9</math> = <math>512</math> subsets(<math>1</math> isn't needed) we have the following:
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<cmath>2^7 + 2^7 + 2^6 + 2^6 = 384</cmath>
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<math>...</math> The total number of sets with at least one of the groups (with repeats)
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For sets that contain two of the groups, we have:
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=<math>2^7 + 2^7 + 2^6 + 2^6</math>
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<cmath>2^5 + 2^5 + 2^5 + 2^5 + 2^4 + 2^4 = 160</cmath>
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= <math>384</math>
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For sets that contain three of the groups, we have:
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<math>...</math> The total number of sets with at least two of the groups (with repeats)
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<cmath>2^4 + 2^3 + 2^3 + 2^3 = 40</cmath>
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= 1st & 2nd pair <math>+</math> 1st & 3rd pair <math>+</math> 1st & 4th pair <math>+</math> 2nd & 3rd pair <math>+</math> 2nd & 4th pair <math>+</math> 3rd & 4th pair
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For sets that contain all of the groups, we have:
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= <math>2^5 + 2^5 + 2^5 + 2^5 + 2^4 + 2^4</math>
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<cmath>2^2 = 4</cmath>
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= <math>160</math>
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By the principle of inclusion and exclusion, the number of product-free subsets is
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<math>...</math> The total number of sets with at least three of the groups (with repeats)
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<cmath>512 - (384 - 160 + 40 - 4) = \boxed{252}</cmath>.
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= 1st, 2nd, & 3rd groups <math>+</math> 1st, 2nd & 4th groups <math>+</math> 1st, 3rd, & 4th groups <math>+</math> 2nd, 3rd, & 4th groups
 
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= <math>2^4 + 2^3 + 2^3 + 2^3</math>
 
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= <math>40</math>
 
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<math>...</math> The total number of sets with all of the groups.
 
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= <math>2^2</math>
 
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= <math>4</math>
 
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<math>...</math> <math>(512-(384-160+40-4)) \implies \boxed{252}</math>.
 
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(Note: If someone can add breaks and maybe help explain why there are powers of 2 that would be good)
 
  
 
==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 16:08, 27 March 2018

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
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All AIME Problems and Solutions

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