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

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== 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)
+
If we start with <math>1,</math> we can include nothing else, so that won't work. By the way, note that <math>1</math> is not an answer choice.
  
 
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.
 
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.
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==Solution 2==
 
==Solution 2==
−
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.
  
−
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>
+
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>
 
The least number is <math>4,</math> so the answer is <math>\boxed{\textbf{(C)}\ 4}.</math>
 
The least number is <math>4,</math> so the answer is <math>\boxed{\textbf{(C)}\ 4}.</math>
  
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Trying <math>4,</math> we can get <math>4,5,6,7,9,11.</math> So <math>4</math> works.
 
Trying <math>4,</math> we can get <math>4,5,6,7,9,11.</math> So <math>4</math> works.
 
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>
 
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,\cdots,12\}</math> into six nonempty subsets such that
  
 
==Video Solution==
 
==Video Solution==

Revision as of 20:45, 27 August 2021

The following problem is from both the 2018 AMC 12A #12 and 2018 AMC 10A #17, 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

If we start with $1,$ we can include nothing else, so that won't work. By the way, note that $1$ is not an answer choice.

If we start with $2,$ 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 $\{1,2,\cdots,12\}$ into six nonempty subsets such that

Video Solution

https://youtu.be/M22S82Am2zM

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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