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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 (Variables)==
−
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 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>
 
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>
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Solution 2==
+
==Solution 2 (Elimination)==
  
−
We can also easily test all the answer choices.  
+
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 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,
+
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 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,
+
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>
  
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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,
+
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 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.
+
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>
 
So, our answer is <math>\boxed{\textbf{(D)}\ 11}.</math>
  
−
==Video Solution 1==
+
==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.

Problem

Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$

Solution 1 (Variables)

Let $M$ be the median. It follows that the two largest integers are both $M+2.$

Let $a$ and $b$ be the two smallest integers such that $a<b.$ The sorted list is \[a,b,M,M+2,M+2.\] Since the median is $2$ greater than their arithmetic mean, we have $\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,$ or \[a+b+14=2M.\] Note that $a+b$ must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let $a=1$ and $b=3,$ from which $M=9$ and $M+2=\boxed{\textbf{(D)}\ 11}.$

~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 $\textbf{(A)},$ 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 $\textbf{(B)},$ 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: $\underline{\hspace{3mm}},\underline{\hspace{3mm}},5,7,7.$ This exceeds the sum of $15.$

Now we try answer choice $\textbf{(C)}.$ 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 $\underline{\hspace{3mm}},\underline{\hspace{3mm}},7,9,9,$ which is exactly $25.$ However, our list needs positive integers, so this won't work.

Since we were really close on answer choice $\textbf{(C)},$ we can intuitively feel that the answer is probably going to be $\textbf{(D)}.$ We can confirm this by creating a list that satisfies the problem and choose $\textbf{(D)}: 1,3,9,11,11.$

So, our answer is $\boxed{\textbf{(D)}\ 11}.$

Video Solution (🚀 Very Fast 🚀)

https://youtu.be/2tx9GEbIRxU

~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

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

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