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

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==Problem 12==
 
==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,..., 7, 8, 9, 10\}</math>.
  
 
==Solution 1 (Casework)==
 
==Solution 1 (Casework)==
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==Solution 2 (PIE)==
 
==Solution 2 (PIE)==
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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:
+
We will consider the <math>2^9 = 512</math> subsets that do not contain 1. 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:
 
<cmath>2^7 + 2^7 + 2^6 + 2^6 = 384</cmath>
 
<cmath>2^7 + 2^7 + 2^6 + 2^6 = 384</cmath>
 
For sets that contain two of the groups, we have:
 
For sets that contain two of the groups, we have:
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<cmath>2^2 = 4</cmath>
 
<cmath>2^2 = 4</cmath>
 
By the principle of inclusion and exclusion, the number of product-free subsets is
 
By the principle of inclusion and exclusion, the number of product-free subsets is
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<cmath>512 - (384 - 160 + 40 - 4) = \boxed{252}</cmath>.
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<cmath>512 - 384 + 160 - 40 + 4 = \boxed{252}</cmath>.
 +
 
 +
==Solution 3==
 +
Let <math>X</math> be a product-free subset, and note that 1 is not in <math>x</math>. We consider four cases:
 +
 
 +
1.) both 2 and 3 are not in <math>X</math>. Then there are <math>2^7=128</math> possible subsets for this case.
 +
 
 +
2.) 2 is in <math>X</math>, but 3 is not. Then 4 in not in <math>X</math>, so there are <math>2^6=64</math> subsets; however, there is a <math>\frac{1}{4}</math> chance that 5 and 10 are both in <math>X</math>, so there are <math>64\cdot \frac{3}{4}=48</math> subsets for this case.
 +
 
 +
3.) 2 is not in <math>X</math>, but 3 is. Then, 9 is not in <math>X</math>, so there are <math>2^6=64</math> subsets for this case.
 +
 
 +
4.) 2 and 3 are both in <math>X</math>. Then, 4, 6, and 9 are not in <math>X</math>, so there are <math>2^4=16</math> total subsets; however, there is a <math>\frac{1}{4}</math> chance that 5 and 10 are both in <math>X</math>, so there are <math>16\cdot \frac{3}{4}=12</math> subsets for this case.
 +
 
 +
Hence our answer is <math>128+48+64+12=\boxed{252}</math>. -Stormersyle
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 +
==Solution 4==
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 +
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 (<math>2^3 = 8</math> 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 <math>4, 6, 9, 10</math> are of type 1. However, <math>8</math> is of type 2. Therefore this case contributes 2.
 +
 
 +
Summing over the 8 cases we get <math>2 + 4 + 8 + 16 + 16 + 16 + 32 + 32 = 126.</math> Multiplying by <math>2</math> because of the <math>7</math> gives <math>252.</math>
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==Solution 5==
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Note that if any element <math>s\geq6</math> makes an invalid set, it must be <math>c</math> of <math>ab=c</math>. Otherwise, <math>ab\geq12>10</math> so no <math>ab=c</math> will suffice. Therefore, whether or not an element <math>s\geq6</math> depends only on the previous elements <math>1\leq\omega\leq5</math>.
 +
 
 +
First, if a set is product-free, it must never contain an element <math>\omega=1</math> or <math>1\cdot1=1</math>.
 +
 
 +
We do casework on the subset of <math>S\in{1,2,3,4,5}=s_1</math> to determine <math>S\in{6,7,8,9,10}=s_2</math> (There are only 12 to cover so we just list them out):
 +
 
 +
When it is empty, there are clearly no restrictions for <math>s_2</math> so there are <math>2^5=32</math> cases.
 +
 
 +
<math>s_1={2}, s_1={4}, s_1={5}</math> has no restrictions <math>s_2</math> so we get <math>2^5\cdot3=96</math> total cases.
 +
 
 +
<math>s_1=3</math> only cannot contain 9 so there are <math>2^4=16</math> cases.
 +
 
 +
<math>s_1={2, 5}; s_1={3, 4}; s_1={3, 5}</math> each have one restriction in <math>s_2</math> so there are <math>2^4\cdot3=48</math> ways.
 +
 
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<math>s_1={2, 3}</math> has 2 restrictions (6 and 9) so there are <math>2^3=8</math> ways.
 +
 
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<math>s_1={4, 5}</math> has no restrictions so there are <math>2^5=32</math> ways.
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<math>s_1={2, 3, 5}</math> has 3 restrictions (6, 9, and 10) so there are <math>2^2=4</math> ways.
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<math>s_1={3, 4, 5}</math> has 1 restriction so there are <math>2^4=16</math> cases.
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 +
Summing the numbers, we get <math>32+96+16+48+8+32+4+16=\boxed{252}</math>.
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/jRZQUv4hY_k?t=504
 +
 
 +
~ pi_is_3.14
  
 
==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}}

Latest revision as of 11:46, 24 January 2024

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 $252.$

Solution 5

Note that if any element $s\geq6$ makes an invalid set, it must be $c$ of $ab=c$. Otherwise, $ab\geq12>10$ so no $ab=c$ will suffice. Therefore, whether or not an element $s\geq6$ depends only on the previous elements $1\leq\omega\leq5$.

First, if a set is product-free, it must never contain an element $\omega=1$ or $1\cdot1=1$.

We do casework on the subset of $S\in{1,2,3,4,5}=s_1$ to determine $S\in{6,7,8,9,10}=s_2$ (There are only 12 to cover so we just list them out):

When it is empty, there are clearly no restrictions for $s_2$ so there are $2^5=32$ cases.

$s_1={2}, s_1={4}, s_1={5}$ has no restrictions $s_2$ so we get $2^5\cdot3=96$ total cases.

$s_1=3$ only cannot contain 9 so there are $2^4=16$ cases.

$s_1={2, 5}; s_1={3, 4}; s_1={3, 5}$ each have one restriction in $s_2$ so there are $2^4\cdot3=48$ ways.

$s_1={2, 3}$ has 2 restrictions (6 and 9) so there are $2^3=8$ ways.

$s_1={4, 5}$ has no restrictions so there are $2^5=32$ ways.

$s_1={2, 3, 5}$ has 3 restrictions (6, 9, and 10) so there are $2^2=4$ ways.

$s_1={3, 4, 5}$ has 1 restriction so there are $2^4=16$ cases.

Summing the numbers, we get $32+96+16+48+8+32+4+16=\boxed{252}$.

Video Solution by OmegaLearn

https://youtu.be/jRZQUv4hY_k?t=504

~ pi_is_3.14

See Also

2017 AIME I (Problems • Answer Key • Resources)
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Problem 11 		Followed by
Problem 13
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