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 1 (Variables)== |
| − | + | Let <math>M</math> be the median. It follows that the two largest integers are both <math>M+2.</math> | |
| + | |||
| + | Let <math>a</math> and <math>b</math> be the two smallest integers such that <math>a<b.</math> The sorted list is <cmath>a,b,M,M+2,M+2.</cmath> | ||
| + | Since the median is <math>2</math> greater than their arithmetic mean, we have <math>\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,</math> or <cmath>a+b+14=2M.</cmath> | ||
| + | Note that <math>a+b</math> must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let <math>a=1</math> and <math>b=3,</math> from which <math>M=9</math> and <math>M+2=\boxed{\textbf{(D)}\ 11}.</math> | ||
| + | |||
| + | ~MRENTHUSIASM | ||
| + | |||
| + | ==Solution 2 (Elimination)== | ||
| − | + | We can also easily test all the answer choices. (This strategy is generally good to use for multiple-choice questions if you don't have a concrete method to proceed with!) | |
| + | For answer choice <math>\textbf{(A)},</math> the mode is <math>5,</math> the median is <math>3,</math> and the arithmetic mean is <math>1.</math> However, we can quickly see this doesn't work, as there are five integers, and they can't have an arithmetic mean of <math>1</math> while having a mode of <math>5.</math> | ||
| + | Trying answer choice <math>\textbf{(B)},</math> the mode is <math>7,</math> the median is <math>5,</math> and the arithmetic mean is <math>3.</math> From the arithmetic mean, we know that all the numbers have to sum to <math>15.</math> We know three of the numbers: <math>\underline{\hspace{3mm}},\underline{\hspace{3mm}},5,7,7.</math> This exceeds the sum of <math>15.</math> | ||
| − | + | Now we try answer choice <math>\textbf{(C)}.</math> The mode is <math>9,</math> the median is <math>7,</math> and the arithmetic mean is <math>5.</math> From the arithmetic mean, we know that the list sums to <math>25.</math> Three of the numbers are <math>\underline{\hspace{3mm}},\underline{\hspace{3mm}},7,9,9,</math> which is exactly <math>25.</math> However, our list needs positive integers, so this won't work. | |
| − | |||
| − | + | Since we were really close on answer choice <math>\textbf{(C)},</math> we can intuitively feel that the answer is probably going to be <math>\textbf{(D)}.</math> We can confirm this by creating a list that satisfies the problem and choose <math>\textbf{(D)}: 1,3,9,11,11.</math> | |
| − | |||
| − | |||
| − | + | So, our answer is <math>\boxed{\textbf{(D)}\ 11}.</math> | |
| − | ==Video Solution | + | ==Video Solution (🚀 Very Fast 🚀)== |
https://youtu.be/2tx9GEbIRxU | https://youtu.be/2tx9GEbIRxU | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution(1-16)== | ||
| + | https://youtu.be/SCwQ9jUfr0g | ||
| + | |||
| + | ~~Hayabusa1 | ||
| + | ==Video Solution by Interstigation== | ||
| + | https://youtu.be/_KNR0JV5rdI?t=1241 | ||
== See Also == | == See Also == | ||
Latest revision as of 23:01, 11 September 2024
- 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 (Variables)
Let 
be the median. It follows that the two largest integers are both 
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 (Elimination)
We can also easily test all the answer choices. (This strategy is generally good to use for multiple-choice questions if you don't have a concrete method to proceed with!)
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 (🚀 Very Fast 🚀)
~Education, the Study of Everything
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. 
