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Difference between revisions of "2018 AMC 10A Problems/Problem 17"
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== Solution ==
 
== Solution ==
 
Intuitively, one would see this list and start with prime numbers. However, there are only 5 prime numbers less than <math>12</math>, making this impossible. It is also clear that another number can't be added in, so <math>2</math> can't be the smallest. Next, we start the sequence with <math>3</math>, and a bit of trial and error shows it's impossible. Lastly, 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>.
 
Intuitively, one would see this list and start with prime numbers. However, there are only 5 prime numbers less than <math>12</math>, making this impossible. It is also clear that another number can't be added in, so <math>2</math> can't be the smallest. Next, we start the sequence with <math>3</math>, and a bit of trial and error shows it's impossible. Lastly, 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>.
 +
(Random_Guy)

Revision as of 17:32, 8 February 2018

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 values 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

Intuitively, one would see this list and start with prime numbers. However, there are only 5 prime numbers less than $12$, making this impossible. It is also clear that another number can't be added in, so $2$ can't be the smallest. Next, we start the sequence with $3$, and a bit of trial and error shows it's impossible. Lastly, starting with $4$, we find that the sequence $4,5,6,7,9,11$ works, giving us $\boxed{\textbf{(C)} \text{ 4}}$. (Random_Guy)

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