Difference between revisions of "2000 AMC 12 Problems/Problem 14"
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We apply casework upon the median: | We apply casework upon the median: | ||
*If the median is <math>2</math> (<math>x \le 2</math>), then the arithmetic progression must be constant. | *If the median is <math>2</math> (<math>x \le 2</math>), then the arithmetic progression must be constant. | ||
− | *If the median is <math>4</math> (<math>x \ge 4</math>), because the mode is <math>2</math>, the mean can either be <math>0,3,6</math> to form an arithmetic progression. Solving for <math>x</math> yields <math>-25,-4,17</math> respectively, of which only <math>17</math> works because it is larger than <math> | + | *If the median is <math>4</math> (<math>x \ge 4</math>), because the mode is <math>2</math>, the mean can either be <math>0,3,6</math> to form an arithmetic progression. Solving for <math>x</math> yields <math>-25,-4,17</math> respectively, of which only <math>17</math> works because it is larger than <math>4</math>. |
*If the median is <math>x</math> (<math>2 \le x \le 4</math>), again, because the mode is <math>2</math>, the mean can either be <math>1, 5/2, 4</math> to form an arithmetic progression. Solving for <math>x</math> yields <math>-18,-7.5, 3</math> respectively, of which only <math>3</math> works because it is greater than <math>2</math>, and less than <math>4</math>. | *If the median is <math>x</math> (<math>2 \le x \le 4</math>), again, because the mode is <math>2</math>, the mean can either be <math>1, 5/2, 4</math> to form an arithmetic progression. Solving for <math>x</math> yields <math>-18,-7.5, 3</math> respectively, of which only <math>3</math> works because it is greater than <math>2</math>, and less than <math>4</math>. | ||
Revision as of 11:20, 24 May 2018
- The following problem is from both the 2000 AMC 12 #14 and 2000 AMC 10 #23, so both problems redirect to this page.
Problem
When the mean, median, and mode of the list
are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ?
Solution
- The mean is .
- Arranged in increasing order, the list is , so the median is either or depending upon the value of .
- The mode is , since it appears three times.
We apply casework upon the median:
- If the median is (), then the arithmetic progression must be constant.
- If the median is (), because the mode is , the mean can either be to form an arithmetic progression. Solving for yields respectively, of which only works because it is larger than .
- If the median is (), again, because the mode is , the mean can either be to form an arithmetic progression. Solving for yields respectively, of which only works because it is greater than , and less than .
The answer is .
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 |
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.