2018 AMC 12A Problem #6

Algebra Level 2

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


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2 solutions

Zain Majumder
Feb 11, 2018

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

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

Since the mean is n n , m + ( m + 4 ) + ( m + 10 ) + ( n + 1 ) + ( n + 2 ) + 2 n 6 = 3 m + 4 n + 17 6 = n 3 m + 17 = 2 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 3 m + 17 = 2 ( m + 11 ) = 2 m + 22 m = 5 3m+17 = 2(m+11) = 2m+22 \implies m=5 .

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

Jerry McKenzie
Feb 9, 2018

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

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

Then it follows:

1 < m 11 < m + 10 < n + 1 10 < n n + 2 < 2 n 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 , 2 n } \{m,m+4,m+10,n+1,n+2,2n\}

Now using the median:

m + n + 11 = 2 n Eqn 2 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.

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