2018 AMC 12A Problems/Problem 6


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 $m+n$?


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

Solution 3

Since the median is $n$, then $\frac{m+10+n+1}{2} = n \Rightarrow m+11 = n$, or $m= n-11$. Plug this in for $m$ values to get $\frac{7n-16}{6} = n \Rightarrow 7n-16 = 6n \Rightarrow n= 16$. Plug it back in to get $m = 5$, thus $16 + 5 = \boxed{21}$.

~ iron

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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

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