2018 AMC 10A Problems/Problem 17

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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