# 2001 AMC 8 Problems/Problem 21

## Problem

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 $\text{(A)}\ 19 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 40$

## Solution

Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is $18$, and there are $2$ numbers less than $18$ and $2$ numbers greater than $18$. The sum of these integers is $5(15)=75$, since the mean is $15$. To make the largest possible number with a given sum, the other numbers must be as small as possible. The two numbers less than $18$ must be positive and distinct, so the smallest possible numbers for these are $1$ and $2$. The number right after $18$ also needs to be as small as possible, so it must be $19$. This means that the remaining number, the maximum possible value for a number in the set, is $75-1-2-18-19=35, \boxed{\text{(D) 35}}$.

## See Also

 2001 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

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