Problem 21 · 2001 AMC 8
Stretch
Arithmetic & Operations
sum-constraintwork-backward
The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is
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Answer: D — 35.
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Hint 1 of 2
The mean turns into a fixed total: five numbers averaging 15 must sum to 75. That total is a budget shared among all five.
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Hint 2 of 2
To make ONE number as big as possible, you have to starve the other four down to the smallest values the rules (distinct, positive, median 18) permit.
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Approach: fix the total, minimize the others
- Convert the mean to a total: the five numbers sum to 5 × 15 = 75. This is the budget — making the largest big means making everyone else small.
- Order them a < b < 18 < d < e. Below the median, the smallest distinct positives are b = 2 and a = 1. Just above the median, d must be a distinct integer > 18, so the smallest allowed is d = 19.
- Whatever's left of the budget goes to e: e = 75 − 1 − 2 − 18 − 19 = 35. (Check: 1, 2, 18, 19, 35 are five distinct positives, median 18, sum 75. ✓) The strategy — fix the total, then minimize everything except your target — maximizes any single value.
- Why must d be 19 and not 18? The integers are different, so d can't tie the median; the next integer up is 19.
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