2022 AMC 10B Problems/Problem 10
- The following problem is from both the 2022 AMC 10B #10 and 2022 AMC 12B #7, so both problems redirect to this page.
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?
Video Solution 1
~Education, the Study of Everything
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 ![\[a,b,M,M+2,M+2.\]](http://latex.artofproblemsolving.com/9/a/0/9a0ec4d708d73a1c4bb1a86ddef45f82d1310e0c.png)
Since the median is greater than their arithmetic mean, we have 
or ![\[a+b+14=2M.\]](http://latex.artofproblemsolving.com/9/b/0/9b0088148fb4116cbdabacd5e4e1c77d86cfd961.png)
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. 
