Difference between revisions of "2002 AMC 12A Problems/Problem 15"
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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 | + | 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 | |
− | + | <math>\boxed{\text{(D)}\ 15 }</math>. | |
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== See Also == | == See Also == |
Latest 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
Solution 1
As the unique mode is , there are at least two s.
As the range is and one of the numbers is , the largest one can be at most .
If the largest one is , then the smallest one is , and thus the mean is strictly larger than , 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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.