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

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 the largest one is $15$, then the smallest one is $7$. This means that we already know four of the values: $8$, $8$, $7$, $15$. Since the mean of all the numbers is $8$, their sum must be $64$. Thus the sum of the missing four numbers is $64-8-8-7-15=26$. But if $7$ is the smallest number, then the sum of the missing numbers must be at least $4\cdot 7=28$, which is again a contradiction.

If the largest number is $14$, we can easily find the solution $(6,6,6,8,8,8,8,14)$. Hence, our answer is $\boxed{\text{(D)}\ 14 }$.

Note

The solution for $14$ is, in fact, unique. As the median must be $8$, this means that both the $4^\text{th}$ and the $5^\text{th}$ number, when ordered by size, must be $8$s. This gives the partial solution $(6,a,b,8,8,c,d,14)$. For the mean to be $8$ each missing variable must be replaced by the smallest allowed value. The solution that works is $(6,6,6,8,8,8,8,14)$

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
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 (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 AMC 10 Problems and Solutions

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