2018 AMC 12A Problem 6

Below is the professionally curated solution for Problem 6 of the 2018 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12A solutions, or check the answer key.

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Concepts:mean median (data)system of equations

Difficulty rating: 1350

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 $m + n?$

$20$

$21$

$22$

$23$

$24$

Solution:

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 $\frac{(n-11)+(n-7)+(n-1)+(n+1)+(n+2)+2n}{6} = n,$ so $7n - 16 = 6n$ and $n = 16.$ Then $m = 5,$ and $m + n = 21.$

Thus, the correct answer is B.

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