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.
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 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 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
Solution 2
We can also easily test all the answer choices.
For answer choice the mode is the median is and the arithmetic mean is However, we can quickly see this doesn't work, as there are five integers, and they can't have an arithmetic mean of while having a mode of
Trying answer choice the mode is the median is and the arithmetic mean is From the arithmetic mean, we know that all the numbers have to sum to We know three of the numbers: This exceeds the sum of
Now we try answer choice The mode is the median is and the arithmetic mean is From the arithmetic mean, we know that the list sums to Three of the numbers are which is exactly However, our list needs positive integers, so this won't work.
Since we were really close on answer choice we can intuitively feel that the answer is probably going to be We can confirm this by creating a list that satisfies the problem and choose
So, our answer is
Video Solution 1
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Video Solution(1-16)
~~Hayabusa1
Video Solution by Interstigation
https://youtu.be/_KNR0JV5rdI?t=1241
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.