Difference between revisions of "2000 AMC 12 Problems/Problem 14"
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| + | {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #14]] and [[2000 AMC 10 Problems|2000 AMC 10 #23]]}} | ||
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== Problem == | == Problem == | ||
When the [[mean]], [[median]], and [[mode]] of the list | When the [[mean]], [[median]], and [[mode]] of the list | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2000|num-b=13|num-a=15}} | {{AMC12 box|year=2000|num-b=13|num-a=15}} | ||
| + | {{AMC10 box|year=2000|num-b=22|num-a=24}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 23:06, 26 November 2011
- 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, which results in a contradiction. - If the median is
(
), then the mean can either be 
to form an arithmetic progression. Solving for yields 
respectively, of which only 
works. - If the median is (
), then the mean can either be 
to form an arithmetic progression. Solving for yields 
respectively, of which only 
works.
), then the arithmetic progression must be constant, which results in a contradiction.
(
), then the mean can either be 
to form an arithmetic progression. Solving for 
respectively, of which only 
works.
), then the mean can either be 
to form an arithmetic progression. Solving for 
respectively, of which only 
works.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 | ||
