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>), | + | *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>), | + | *If the median is <math>x</math> (<math>2 \le x \le 4</math>), we must have the arithmetic progression <math>2, x, \frac{25+x}{7}</math>. Thus, we find that <math>2x=2+\frac{25+x}{7}</math> so <math>x=3</math>. |
| − | The answer is <math>3 + 17 = 20\ \mathrm{(E)}</math>. | + | The answer is <math>3 + 17 = \boxed{20\ \mathrm{(E)}}</math>. |
| + | -Error fixed(BIGFOOT09) Changes Copyrighted© | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/IziHKOubUI8?t=933 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | == Video Solution == | ||
| + | https://youtu.be/OpgSamhXPTg | ||
| + | |||
| + | https://www.youtube.com/watch?v=fjOc_c7rpy0 ~David | ||
== See also == | == See also == | ||
Latest revision as of 00:12, 10 November 2024
- 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
![\[10,2,5,2,4,2,x\]](http://latex.artofproblemsolving.com/1/5/a/15a3e937aa574fc69dcbb923898a66dfcfe44ee3.png)
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.
.
, so the median is either 
or 
, 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 (
), we must have the arithmetic progression 
. Thus, we find that 
so 
.
), then the arithmetic progression must be constant.
(
), because the mode is 
to form an arithmetic progression. Solving for 
respectively, of which only 
works because it is larger than 
), we must have the arithmetic progression 
. Thus, we find that 
so 
.The answer is 
.
-Error fixed(BIGFOOT09) Changes Copyrighted©
Video Solution by OmegaLearn
https://youtu.be/IziHKOubUI8?t=933
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=fjOc_c7rpy0 ~David
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. 
