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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)
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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>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.
  
−
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.
+
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.
  
−
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>.
+
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>
  
 
==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.
 
We know that all the 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> 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)} \text{ 4}}</math>.
+
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*}
−
<math>1:</math> <math>\text{all}</math> <math>\text{numbers}</math> <math>\text{within}</math> <math>\text{range}</math>
+
1&: \ \text{all elements of }\{1,2,\dots,12\} \\
−
 
+
2&: \ 4,6,8,10,12 \\
−
<math>2: 4,6,8,10,12</math>
+
3&: \ 6,9,12 \\
−
 
+
4&: \ 8,12 \\
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<math>3: 6,9,12</math>
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5&: \ 10 \\
−
 
+
6&: \ 12
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<math>4: 8,12</math>
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\end{align*}</cmath>
−
 
 
−
<math>5: 10</math>
 
−
 
 
−
<math>6: 12</math>
 
−
 
 
 
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.
 
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 <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> 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>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 16:47, 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. (Also 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 the odd numbers except $1,$ namely $3, 5, 7, 9, 11,$ can be used.

Now we have $7$ 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 $\boxed{\textbf{(C)}\ 4}.$

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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