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

(Created page with "Let <math>S</math> be a set of 6 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> w...")
 
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== Problem ==
 
Let <math>S</math> be a set of 6 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 values of an element in <math>S?</math>
 
Let <math>S</math> be a set of 6 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 values 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>
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== Solution ==
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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>.

Revision as of 18:30, 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 $\boxed{\textbf{(C)} \text{ 4}}$.

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