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Difference between revisions of "2018 AMC 10A Problems/Problem 17"

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{{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #12]] and [[2018 AMC 10A Problems|2018 AMC 10A #17]]}}
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{{duplicate|[[2018 AMC 10A Problems/Problem 17|2018 AMC 10A #17]] and [[2018 AMC 12A Problems/Problem 12|2018 AMC 12A #12]]}}
  
 
== Problem ==
 
== Problem ==
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Let <math>S</math> be a set of <math>6</math> integers taken from <math>\{1,2,\dots,12\}</math> with the property that if <math>a</math> and <math>b</math> are elements of <math>S</math> with <math>a<b</math>, then <math>b</math> is not a multiple of <math>a</math>. What is the least possible value of an element in <math>S?</math>
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Let <math>S</math> be a set of <math>6</math> integers taken from <math>\{1,2,\dots,12\}</math> with the property that if <math>a</math> and <math>b</math> are elements of <math>S</math> with <math>a<b</math>, then <math>b</math> is not a multiple of <math>a</math>. What is the least possible value of an element in <math>S</math>?
  
 
<math>\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7</math>
 
<math>\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7</math>
  
 
== Solution 1 ==
 
== Solution 1 ==
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If we start with <math>1</math>, we can include nothing else, so that won't work. (Also note that <math>1</math> is not an answer choice)
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We start with <math>2</math> because <math>1</math> is not an answer choice. We would have to include every odd number except <math>1</math> to fill out the set, but then <math>3</math> and <math>9</math> would violate the rule, so that won't work.
  
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If we start with <math>2</math>, we would have to include every odd number except <math>1</math> to fill out the set, but then <math>3</math> and <math>9</math> would violate the rule, so that won't work.
+
Experimentation with <math>3</math> shows it's likewise impossible. You can include <math>7,11,</math> and either <math>5</math> or <math>10</math> (which are always safe). But after adding either <math>4</math> or <math>8</math> we have no more valid numbers.
  
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Experimentation with <math>3</math> shows it's likewise impossible. You can include <math>7</math>, <math>11</math>, and either <math>5</math> or <math>10</math> (which are always safe). But after adding either <math>4</math> or <math>8</math> we have no more valid numbers.
+
Finally, starting with <math>4,</math> we find that the sequence <math>4,5,6,7,9,11</math> works, giving us <math>\boxed{\textbf{(C)}\ 4}.</math>
−
 
 
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Finally, starting with <math>4</math>, we find that the sequence <math>4,5,6,7,9,11</math> works, giving us <math>\boxed{\textbf{(C)} \text{ 4}}</math>.
 
  
 
==Solution 2==
 
==Solution 2==
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We know that all the odd numbers except <math>1,</math> namely <math>3, 5, 7, 9, 11,</math> can be used.
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We know that all odd numbers except <math>1,</math> namely <math>3, 5, 7, 9, 11,</math> can be used.
  
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Now we have <math>7</math> to choose from for the last number (out of <math>1, 2, 4, 6, 8, 10, 12</math>). We can eliminate <math>1, 2, 10</math>, and <math>12,</math> and we have <math>4, 6, 8</math> to choose from. However, <math>9</math> is a multiple of <math>3.</math> Now we have to take out either <math>3</math> or <math>9</math> from the list. If we take out <math>9,</math> none of the numbers would work, but if we take out <math>3</math>, we get <cmath>4, 5, 6, 7, 9, 11.</cmath>
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Now we have <math>7</math> possibilities to choose from for the last number (out of <math>1, 2, 4, 6, 8, 10, 12</math>). We can eliminate <math>1, 2, 10,</math> and <math>12,</math> and we have <math>4, 6, 8</math> to choose from. However, <math>9</math> is a multiple of <math>3.</math> Now we have to take out either <math>3</math> or <math>9</math> from the list. If we take out <math>9,</math> none of the numbers would work, but if we take out <math>3,</math> we get <cmath>4, 5, 6, 7, 9, 11.</cmath>
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The least number is <math>4</math>, so the answer is <math>\boxed{\textbf{(C)} \text{ 4}}</math>.
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The least number is <math>4,</math> so the answer is <math>\boxed{\textbf{(C)}\ 4}.</math>
  
 
==Solution 3==
 
==Solution 3==
 
We can get the multiples for the numbers in the original set with multiples in the same original set
 
We can get the multiples for the numbers in the original set with multiples in the same original set
 +
<cmath>\begin{align*}
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1&: \ \text{all elements of }\{1,2,\dots,12\} \\
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2&: \ 4,6,8,10,12 \\
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3&: \ 6,9,12 \\
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4&: \ 8,12 \\
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5&: \ 10 \\
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6&: \ 12
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\end{align*}</cmath>
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It will be safe to start with <math>5</math> or <math>6</math> since they have the smallest number of multiples as listed above, but since the question asks for the least, it will be better to try others.
 +
 +
Trying <math>4,</math> we can get <math>4,5,6,7,9,11.</math> So <math>4</math> works.
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Trying <math>3,</math> it won't work, so the least is <math>4.</math> This means the answer is <math>\boxed{\textbf{(C)}\ 4}.</math>
 +
 +
==Solution 4==
 +
We partition <math>\{1,2,\ldots,12\}</math> into six nonempty subsets such that for every subset, each element is a multiple of all elements less than or equal to itself: <cmath>\{1,2,4,8\}, \ \{3,6,12\}, \ \{5,10\}, \ \{7\}, \ \{9\}, \ \{11\}.</cmath>
 +
Clearly, <math>S</math> must contain exactly one element from each subset:
 +
 +
* For <math>\{5,10\},</math> we can select either <math>5</math> or <math>10.</math>
  
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<math>1:</math> <math>\text{all}</math> <math>\text{numbers}</math> <math>\text{within}</math> <math>\text{range}</math>
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* For <math>\{3,6,12\},</math> we can select either <math>6</math> or <math>12.</math> Recall that since <math>9\in S,</math> we cannot select <math>3.</math>
  
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<math>2: 4,6,8,10,12</math>
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* For <math>\{1,2,4,8\},</math> we can select either <math>4</math> (provided that <math>12\not\in S</math>) or <math>8.</math> Recall that since either <math>6\in S</math> or <math>12\in S,</math> we can select neither <math>1</math> nor <math>2.</math>
  
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<math>3: 6,9,12</math>
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If <math>4\in S,</math> then the possibilities of <math>S</math> are <math>\{4,5,6,7,9,11\}</math> or <math>\{4,6,7,9,10,11\}.</math> So, the least possible value of an element in <math>S</math> is <math>\boxed{\textbf{(C)}\ 4}.</math>
  
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<math>4: 8,12</math>
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<u><b>Remark</b></u>
  
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<math>5: 10</math>
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There exist multiple such partitions of <math>\{1,2,\ldots,12\}</math> into six nonempty subsets, one of which is <cmath>\{1,2,6,12\}, \ \{3,9\}, \ \{4,8\}, \ \{5,10\}, \ \{7\}, \ \{11\}.</cmath>
 +
Regardless of which partition we use, we will conclude that to minimize the least element of <math>S,</math> the only possibilities for <math>S</math> are <math>\{4,5,6,7,9,11\}</math> or <math>\{4,6,7,9,10,11\}.</math>
  
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<math>6: 12</math>
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~MRENTHUSIASM
  
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It will be safe to start with 5 or 6 since they have the smallest number of multiples as listed above, but since the question asks for the least, it will be better to try others.
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==Solution 5==
  
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Trying <math>4</math>, we can get <math>4,5,6,7,9,11</math>. So <math>4</math> works.
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We start with 2 as 1 is not an answer option. Our set would be <math>\{2,3,5,7,11\}</math>. We realize we cannot add 12 to the set because 12 is a multiple of 3. Our set only has 5 elements, so starting with 2 won't work.
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Trying <math>3</math> won't work, so the least is <math>4</math>. This means the answer is <math>\boxed{\textbf{(C) } 4}</math>
+
 
 +
We try 3 next. Our set becomes <math>\{3,4,5,7,11\}</math>. We run into the same issue as before so starting with 3 won't work.
 +
 
 +
We then try 4. Our set becomes <math>\{4,5,6,7,9,11\}</math>. We see we have 6 elements with none being multiples of each other. Therefore our answer is <math>\boxed{\textbf{(C)}\ 4}.</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:South South]
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/M22S82Am2zM
 
https://youtu.be/M22S82Am2zM
 +
 +
~IceMatrix
 +
 +
==Video Solution==
 +
https://youtu.be/QtqBEZAgXpk
 +
 +
~savannahsolver
  
 
==See Also ==
 
==See Also ==

Latest revision as of 02:29, 14 October 2023

The following problem is from both the 2018 AMC 10A #17 and 2018 AMC 12A #12, so both problems redirect to this page.

Problem

Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$, then $b$ is not a multiple of $a$. What is the least possible value of an element in $S$?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$

Solution 1

We start with $2$ because $1$ is not an answer choice. We would have to include every odd number except $1$ to fill out the set, but then $3$ and $9$ would violate the rule, so that won't work.

Experimentation with $3$ shows it's likewise impossible. You can include $7,11,$ and either $5$ or $10$ (which are always safe). But after adding either $4$ or $8$ we have no more valid numbers.

Finally, starting with $4,$ we find that the sequence $4,5,6,7,9,11$ works, giving us $\boxed{\textbf{(C)}\ 4}.$

Solution 2

We know that all odd numbers except $1,$ namely $3, 5, 7, 9, 11,$ can be used.

Now we have $7$ possibilities to choose from for the last number (out of $1, 2, 4, 6, 8, 10, 12$). We can eliminate $1, 2, 10,$ and $12,$ and we have $4, 6, 8$ to choose from. However, $9$ is a multiple of $3.$ Now we have to take out either $3$ or $9$ from the list. If we take out $9,$ none of the numbers would work, but if we take out $3,$ we get \[4, 5, 6, 7, 9, 11.\] The least number is $4,$ so the answer is $\boxed{\textbf{(C)}\ 4}.$

Solution 3

We can get the multiples for the numbers in the original set with multiples in the same original set \begin{align*} 1&: \ \text{all elements of }\{1,2,\dots,12\} \\ 2&: \ 4,6,8,10,12 \\ 3&: \ 6,9,12 \\ 4&: \ 8,12 \\ 5&: \ 10 \\ 6&: \ 12 \end{align*} It will be safe to start with $5$ or $6$ since they have the smallest number of multiples as listed above, but since the question asks for the least, it will be better to try others.

Trying $4,$ we can get $4,5,6,7,9,11.$ So $4$ works. Trying $3,$ it won't work, so the least is $4.$ This means the answer is $\boxed{\textbf{(C)}\ 4}.$

Solution 4

We partition $\{1,2,\ldots,12\}$ into six nonempty subsets such that for every subset, each element is a multiple of all elements less than or equal to itself: \[\{1,2,4,8\}, \ \{3,6,12\}, \ \{5,10\}, \ \{7\}, \ \{9\}, \ \{11\}.\] Clearly, $S$ must contain exactly one element from each subset:

  • For $\{5,10\},$ we can select either $5$ or $10.$
  • For $\{3,6,12\},$ we can select either $6$ or $12.$ Recall that since $9\in S,$ we cannot select $3.$
  • For $\{1,2,4,8\},$ we can select either $4$ (provided that $12\not\in S$) or $8.$ Recall that since either $6\in S$ or $12\in S,$ we can select neither $1$ nor $2.$

If $4\in S,$ then the possibilities of $S$ are $\{4,5,6,7,9,11\}$ or $\{4,6,7,9,10,11\}.$ So, the least possible value of an element in $S$ is $\boxed{\textbf{(C)}\ 4}.$

Remark

There exist multiple such partitions of $\{1,2,\ldots,12\}$ into six nonempty subsets, one of which is \[\{1,2,6,12\}, \ \{3,9\}, \ \{4,8\}, \ \{5,10\}, \ \{7\}, \ \{11\}.\] Regardless of which partition we use, we will conclude that to minimize the least element of $S,$ the only possibilities for $S$ are $\{4,5,6,7,9,11\}$ or $\{4,6,7,9,10,11\}.$

~MRENTHUSIASM

Solution 5

We start with 2 as 1 is not an answer option. Our set would be $\{2,3,5,7,11\}$. We realize we cannot add 12 to the set because 12 is a multiple of 3. Our set only has 5 elements, so starting with 2 won't work.

We try 3 next. Our set becomes $\{3,4,5,7,11\}$. We run into the same issue as before so starting with 3 won't work.

We then try 4. Our set becomes $\{4,5,6,7,9,11\}$. We see we have 6 elements with none being multiples of each other. Therefore our answer is $\boxed{\textbf{(C)}\ 4}.$

~South

Video Solution

https://youtu.be/M22S82Am2zM

~IceMatrix

Video Solution

https://youtu.be/QtqBEZAgXpk

~savannahsolver

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2018 AMC 12A (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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