2001 AMC 8 Problem 21

Attempt Problem 21 of the 2001 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

The mean of a set of five different positive integers is 15.15. The median is 18.18. The maximum possible value of the largest of these five integers is
1919
2424
3232
3535
4040
Answer: D
Concepts:meanmedian (data)optimization
Small hint:
The five numbers sum to 7575
Big hint:
Make the other four numbers as small as possible
Solution:
The median of the set of numbers is the third largest number, which is 18.18. There are two numbers less than 1818 and two numbers greater than it.
The mean of the set is 15,15, so the sum of all the numbers is 5⋅15=75.5 \cdot 15 = 75. In order to maximize the largest number with this sum, the other numbers must be as small as possible.
The two numbers less than 1818 must be positive and distinct, so they must be 11 and 2.2.
The number immediately after 1818 must also be as small as possible, so it must be 19.19.
Therefore, the remaining number, the maximum possible value in the set, is 75−1−2−18−19=35. 75 - 1 - 2 - 18 - 19 = 35.
Thus, D is the correct answer.
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