Difference between revisions of "2022 AMC 10B Problems/Problem 10"
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<math>\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13</math> | <math>\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Let <math>M</math> be the median. It follows that the two largest integers are <math>M+2.</math> | Let <math>M</math> be the median. It follows that the two largest integers are <math>M+2.</math> | ||
Revision as of 07:46, 28 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.
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 1
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
Solution 2
We can also easily test all the answer choices.
For answer choice A, the mode is 5, the median is 3, and the arithmetic mean is 1. However, we can quickly see this doesn't work, as there are five integers, and they can't have an arithmetic mean of 1 while having a mode of 5,
Trying answer choice B, the mode is 7, the median is 5, and the arithmetic mean is 3. From the arithmetic mean, we know that all the numbers have to sum to 15. We know three of the numbers: _,_,5,7,7. This exceeds the sum of 15,
Now we try answer choice C. The mode is 9, the median is 7, and the arithmetic mean is 5. From the arithmetic mean, we know that the list sums to 25. Three of the numbers are _,_,7,9,9, which is exactly 25. However, our list needs positive integers, so this won't work,
Since we were really close on choice C, we can intuitively feel that the answer is probably going to be D. We can confirm this by creating a list that satisfies the problem and choice D: 1,3,9,11,11.
So, our answer is 
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. 
