For positive integers m and n such that m + 1 0 < n + 1 , both the mean and the median of the set { m , m + 4 , m + 1 0 , n + 1 , n + 2 , 2 n } are equal to n . What is m + n ?
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Using the average: 3 m + 4 n + 1 7 = 6 n ⇒ 3 m + 1 7 = 2 n Eqn 1
⇒ 2 3 m + 1 9 = n + 1 ⇒ 2 3 m + 1 9 > m + 1 0 ⇒ m > 1
Then it follows:
1 < m ⇒ 1 1 < m + 1 0 < n + 1 ⇒ 1 0 < n ⇒ n + 2 < 2 n
Thus the set is ordered (from least to greatest):
{ m , m + 4 , m + 1 0 , n + 1 , n + 2 , 2 n }
Now using the median:
m + n + 1 1 = 2 n 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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Based on the inequality, the terms are already in order from least to greatest.
Since the median is n , 2 ( m + 1 0 ) + ( n + 1 ) = 2 m + n + 1 1 = n ⟹ m + 1 1 = n .
Since the mean is n , 6 m + ( m + 4 ) + ( m + 1 0 ) + ( n + 1 ) + ( n + 2 ) + 2 n = 6 3 m + 4 n + 1 7 = n ⟹ 3 m + 1 7 = 2 n .
Substituting, we get 3 m + 1 7 = 2 ( m + 1 1 ) = 2 m + 2 2 ⟹ m = 5 .
n = 5 + 1 1 = 1 6 , so our final answer is 5 + 1 6 = 2 1 .