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Difference between revisions of "2002 AMC 12A Problems/Problem 15"

(→‎Solution 1)
(→‎Solution 1)
Line 27: Line 27:
 
If we have 2 8's we can add find the numbers 1, 2, 3, 8, 8, 13, 14, 15.
 
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.
 
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
+
We can also see that they satisfy the need for the mode and median to be 8. This means that the answer will be
 
<math>\boxed{\text{(D)}\ 15 }</math>.
 
<math>\boxed{\text{(D)}\ 15 }</math>.
  

Revision as of 23:21, 13 January 2022

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 $\boxed{\text{(D)}\ 15 }$.

See Also

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

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