2018 AMC 12A Problems/Problem 6

Problem

For positive integers $m$ and $n$ such that $m+10<n+1$ , both the mean and the median of the set $\{m, m+4, m+10, n+1, n+2, 2n\}$ are equal to $n$ . What is $m+n$ ?

$\textbf{(A)}20\qquad\textbf{(B)}21\qquad\textbf{(C)}22\qquad\textbf{(D)}23\qquad\textbf{(E)}24$

Solution 1

The mean and median are $\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,$ so $3m+17=2n$ and $m+11=n$ . Solving this gives $\left(m,n\right)=\left(5,16\right)$ for $m+n=\boxed{21}$ . (trumpeter)

Solution 2

This is an alternate solution if you don't want to solve using algebra. First, notice that the median $n$ is the average of $m+10$ and $n+1$ . Therefore, $n=m+11$ , so the answer is $m+n=2m+11$ , which must be odd. This leaves two remaining options: ${(B) 21}$ and ${(D) 23}$ . Notice that if the answer is $(B)$ , then $m$ is odd, while $m$ is even if the answer is $(D)$ . Since the average of the set is an integer $n$ , the sum of the terms must be even. $4+10+1+2+2n$ is odd by definition, so we know that $3m+2n$ must also be odd, thus with a few simple calculations $m$ is odd. Because all other answers have been eliminated, $(B)$ is the only possibility left. Therefore, $m+n=\boxed{21}$ . ∎ --anna0kear

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2018 AMC 12A Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

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