2002 AMC 12A Problems/Problem 15

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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 $\boxed{\text{(D) }14}$

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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