Difference between revisions of "2000 AMC 12 Problems/Problem 14"
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The answer is <math>3 + 17 = 20\ \mathrm{(E)}</math>. | The answer is <math>3 + 17 = 20\ \mathrm{(E)}</math>. | ||
+ | |||
+ | == 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 18:18, 22 April 2023
- 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
(
), we must have the arithmetic progression
. Thus, we find that
so
.
The answer is .
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.