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.
+6248 \(\text{the numbers are all distinct so they can be ordered}\\ \text{the median will be the average of the 5th and 6th largest numbers}\\ \text{the 5th number must be at least 5 and at most 9}\\ \text{our minimal average sequence is thus}\\ 1,2,3,4,n,20-n,20-n+1,20-n+2, \dots 20-n+4 \\ \text{with mean}\\ \dfrac{10+20+80-4n+10}{10} = \dfrac{120-4n}{10} = 12 - \dfrac 2 5 n\\ \text{This is clearly minimized by choosing }n \text{ as large as possible}\\ \text{The largest possible }n \text{ is 9, and thus the minimum average is }\\ 12 - \dfrac{18}{5} = \dfrac{42}{5}\)
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