Difference between revisions of "2022 AMC 10B Problems/Problem 10"
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==Solution== | ==Solution== |
Revision as of 20:48, 19 November 2022
- The following problem is from both the 2022 AMC 10B #10 and 2022 AMC 12B #7, so both problems redirect to this page.
Contents
Problem
Camila writes down five positive integers. The unique mode of these integers is greater than their median, and the median is
greater than their arithmetic mean. What is the least possible value for the mode?
Solution
Let be the median. It follows that the two largest integers are
Let and
be the two smallest integers such that
The sorted list is
Since the median is
greater than their arithmetic mean, we have
or
Note that
must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let
and
from which
and
~MRENTHUSIASM
Video Solution 1
~Education, the Study of Everything
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.