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 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 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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