2002 AMC 12A Problems/Problem 15

The following problem is from both the 2002 AMC 12A #15 and 2002 AMC 10A #21, so both problems redirect to this page.

Problem

The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is

$\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15$

Solution 1

As the unique mode is $8$ , there are at least two $8$ s.

As the range is $8$ and one of the numbers is $8$ , the largest one can be at most $16$ .

If the largest one is $16$ , then the smallest one is $8$ , and thus the mean is strictly larger than $8$ , which is a contradiction.

If we have 2 8's we can add find the numbers 1, 2, 3, 8, 8, 13, 14, 15. They add to 64 and 64/8 = 8. We can also see that they satisfy the need for the mode and median to be 8. This means that the answer will be 15 $\boxed{\text{(D)}\ 15 }$ .

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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 AMC 12 Problems and Solutions

2002 AMC 10A (Problems • Answer Key • Resources)
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 AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2002 AMC 12A Problems/Problem 15

Problem

Solution 1

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