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 4, 6, 7, 8, 8, 9, 10, 12. This is a possible solution but has not reached the maximum.

If we have 4 8's we can find the numbers 6, 6, 6, 8, 8, 8, 8, 14.

We can also see that they satisfy the need for the mode, median, and range to be 8. This means that the answer will be $\boxed{\text{(D)}\ 14 }$ . ~By QWERTYUIOPASDFGHJKLZXCVBNM

Solution 2

We could express this collection as integers $a\textsubscript{1}$ through $a\textsubscript{8}$ , with $a\textsubscript{1}$ being the smallest and $a\textsubscript{8}$ being the largest.

Since the mean is $8$ , we know that $a\textsubscript{4}$ and $a\textsubscript{5}$ must also be $8$ . If they were not, the other numbers, which are lesser and greater than $a\textsubscript{4}$ and $a\textsubscript{5}$ respectively, would not be able to satisfy the condition that $8$ is the mode.

There are $8$ terms and the mean is $8$ . This tells us that the sum of all the numbers is $64$ .

We want to maximize the value of $a\textsubscript{8}$ , so we set $a\textsubscript{6}$ and $a\textsubscript{7}$ to $8$ as well.

Knowing that we want to minimize numbers and that the range is $8$ , we set $a\textsubscript{1}$ , $a\textsubscript{2}$ , and $a\textsubscript{3}$ equal to $a\textsubscript{8} - 8$ .

{ $a\textsubscript{1}$ , $a\textsubscript{2}$ , $a\textsubscript{3}$ , $a\textsubscript{4}$ , $a\textsubscript{5}$ , $a\textsubscript{6}$ , $a\textsubscript{7}$ , $a\textsubscript{8}$ } $=$ { $a\textsubscript{8} - 8$ , $a\textsubscript{8} - 8$ , $a\textsubscript{8} - 8$ , $8$ , $8$ , $8$ , $8$ , $a\textsubscript{8}$ }

Letting the sum of all the numbers be $64$ , we find that $32 + 4a_8 - 24 = 64$ , which simplifies to $4a_8 = 56$ . Solving, we get $\boxed{\text{(D)}\ 14 }$ . ~By SK80, mod_x for minor edits

Video Solution by OmegaLearn

https://youtu.be/xqo0PgH-h8Y?t=848

~ pi_is_3.14

Video Solution

https://youtu.be/BNa1Ffu517I

https://www.youtube.com/watch?v=Sic_pM_cmrA ~David

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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