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 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 $(6,6,6,8,8,8,8,14)$

Solution 2

Let the 8 numbers be $a, b, c, d, e, f, g, h$ , arranged in increasing order. Since the range of the eight numbers is 8, $h=a+8$ .

I claim that $d$ , $e$ must both be $8$ . Since the median is 8, the mean of $d$ and $e$ must be 8. Let's assume that $d$ and $e$ aren't $8$ . The mode of the collection is $8$ , and if $d$ and $e$ aren't, then $8$ must be between $d$ and $e$ (i.e. not in the collection). This is a contradiction, so $d$ and $e$ have to be 8.

Now, we have the eight numbers are $a, b, c, 8, 8, f, g, a+8$ .

Since the mean is $8$ , we have $a+b+c+8+8+f+g+a+8=8 \times 8 = 64$ , giving us $2a+b+c+f+g=40$ .

Since $f, g \ge 8$ , $f+g \ge 16$ . Plugging that in, we have $2a+b+c \le 24$ . Note that we can't do the same for $b$ and $c$ , but we can do $b+c \ge 2a$ , giving us $4a \le 24$ , which means $a\le 6$ . We want to find $h=a+8\le 6+8=14$ This is our answer, so $\boxed{(D)14}$

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.

Page

Toolbox

Search

2002 AMC 12A Problems/Problem 15

Contents

Problem

Solution 1

Note

Solution 2

See Also