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Difference between revisions of "2022 AMC 10B Problems/Problem 10"
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{{duplicate|[[2022 AMC 10B Problems/Problem 10|2022 AMC 10B #10]] and [[2022 AMC 12B Problems/Problem 7|2022 AMC 12B #7]]}}
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==Problem==
 
==Problem==
 
Camila writes down five positive integers. The unique mode of these integers is <math>2</math> greater than their median, and the median is <math>2</math> greater than their arithmetic mean. What is the least possible value for the mode?
 
Camila writes down five positive integers. The unique mode of these integers is <math>2</math> greater than their median, and the median is <math>2</math> greater than their arithmetic mean. What is the least possible value for the mode?
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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>
  
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==Solution==
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==Solution 1 (Variables)==
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Let <math>M</math> be the median. It follows that the two largest integers are <math>M+2.</math>
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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
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==Solution 2 (Elimination)==
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 +
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!)
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 +
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>
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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 <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.
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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>
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So, our answer is <math>\boxed{\textbf{(D)}\ 11}.</math>
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==Video Solution (🚀 Very Fast 🚀)==
 +
https://youtu.be/2tx9GEbIRxU
 +
 +
~Education, the Study of Everything
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 +
==Video Solution(1-16)==
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https://youtu.be/SCwQ9jUfr0g
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 +
~~Hayabusa1
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==Video Solution by Interstigation==
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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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