2018 AMC 12A Problem #6

Based on the inequality, the terms are already in order from least to greatest.

Since the median is $n$ , $\frac{(m+10)+(n+1)}{2} = \frac{m+n+11}{2} = n \implies m+11 = n$ .

Since the mean is $n$ , $\frac{m+(m+4)+(m+10)+(n+1)+(n+2)+2n}{6} = \frac{3m+4n+17}{6} = n \implies 3m+17 = 2n$ .

Substituting, we get $3m+17 = 2(m+11) = 2m+22 \implies m=5$ .

$n = 5+11 = 16$ , so our final answer is $5+16=\boxed{21}$ .

Using the average: $3m+4n+17=6n\Rightarrow 3m+17 = 2n \qquad \textbf{Eqn 1}$

$\Rightarrow \frac{3m+19}{2}=n+1\Rightarrow\frac{3m+19}{2}>m+10 \Rightarrow m>1$

Then it follows:

$1<m \Rightarrow 11<m+10<n+1 \Rightarrow 10<n \Rightarrow n+2<2n$

Thus the set is ordered (from least to greatest):

$\{m,m+4,m+10,n+1,n+2,2n\}$

Now using the median:

$m+n+11=2n \qquad \textbf{Eqn 2}$

Solving equation 1 and 2, we get m=5 and n=16, thus the required answer is 5+16=21.