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

(Solution 1)
(Solution 1)
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If the largest one is <math>16</math>, then the smallest one is <math>8</math>, and thus the mean is strictly larger than <math>8</math>, which is a contradiction.
 
If the largest one is <math>16</math>, then the smallest one is <math>8</math>, and thus the mean is strictly larger than <math>8</math>, which is a contradiction.
  
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 4, 6, 7, 8, 8, 9, 10, 12.
They add to 64 and 64/8 = 8.
+
This is a possible solution but has not reached the maximum.
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>.
+
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
 +
<math>\boxed{\text{(D)}\ 14 }</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 07:18, 25 February 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 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 $\boxed{\text{(D)}\ 14 }$.

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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