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

You can immediately notice that the median $n$ is the average of $m+10$ and $n+1$ . There fore, $n=m+11$ , so now we know we just are looking for $m+n=2m+11$ , which must be odd. This leaves our two remaining options, {(B)}21\qquad\textbf and {(D)}23\qquad\textbf. Note that if the answer is $(B)$ , then $m$ is odd, while the opposite is true for $m$ if we get $(D)$ . This fact will come in handy later on and prove itself useful when we solve for its parity. Since the average of the set of six numbers $n$ is an integer, 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 some simple calculations $m$ is odd. As with the previous few observations, we have eliminated all other answers, and $(B)$ is the only remaining possibility left. Therefore $m+n=(B)\boxed{21}$ .

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

See Also