Five distinct positive integers represent monthly retention counts, and they sum to 60. What is the greatest possible value of their median?
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Answer & Explanation
Correct answer
C
To make the median (the 3rd of five sorted distinct positive integers) as large as possible, push the other four to be as small as the constraints allow. The two below the median should be the smallest distinct positive integers, 1 and 2. The two above must still be distinct and larger than the median, so the smallest they can be is its closest neighbors, median + 1 and median + 2.
Let the median be m. Then 1 + 2 + m + (m + 1) + (m + 2) = 3m + 6 = 60, so m = 18, realized by {1, 2, 18, 19, 20}. The easily missed move is that the two values above the median must be kept as small as possible (m + 1 and m + 2) rather than enlarged, because every unit added to them is stolen from the sum the median could have used. Checking m = 19 forces a minimum total of 1 + 2 + 19 + 20 + 21 = 63, which exceeds 60, so 18 is the maximum.
A student cannot back-solve a choice without rebuilding the full five-integer set, which is the forward solution itself. Choice 12 is just the mean. Choice 35 maximizes one value while ignoring rank. Choices 16 and 19 come from an arithmetic slip and a distinctness slip respectively.