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Suppose ten distinct, positive integers have a median of 10 . ("Distinct integers" means that no two integers are the same.)

What is the smallest the average of those ten integers could be?

Write your explanatiion in complete sentences.

Jan 31, 2019

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$$\text{the numbers are all distinct so they can be ordered}\\ \text{the median will be the average of the 5th and 6th largest numbers}\\ \text{the 5th number must be at least 5 and at most 9}\\ \text{our minimal average sequence is thus}\\ 1,2,3,4,n,20-n,20-n+1,20-n+2, \dots 20-n+4 \\ \text{with mean}\\ \dfrac{10+20+80-4n+10}{10} = \dfrac{120-4n}{10} = 12 - \dfrac 2 5 n\\ \text{This is clearly minimized by choosing }n \text{ as large as possible}\\ \text{The largest possible }n \text{ is 9, and thus the minimum average is }\\ 12 - \dfrac{18}{5} = \dfrac{42}{5}$$

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Jan 31, 2019