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?

2018 AMC 12A Problem 6

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For positive integers

m

and

n

such that

m + 10 \lt 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?

20

21

22

23

24

The condition

m + 10 \lt n + 1

forces the six values to be listed in increasing order

Set the median

\frac{(m+10)+(n+1)}{2} = n

and set the mean equal to

n,

then solve the two equations

Because

m + 10 \lt n + 1,

the six numbers are already increasing, so the median is the average of the middle two:

\frac{(m+10)+(n+1)}{2} = n,

giving

m = n - 11.

The mean condition is

\tiny \frac{(n-11)+(n-7)+(n-1)+(n+1)+(n+2)+2n}{6} = n,

7n - 16 = 6n

and

n = 16.

Then

m = 5,

and

m + n = 21.

Thus, the correct answer is B.