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Difference between revisions of "2018 AMC 12A Problems/Problem 6"

(Created page with "== Problem == For positive integers <math>m</math> and <math>n</math> such that <math>m+10<n+1</math>, both the mean and the median of the set <math>\{m, m+4, m+10, n+1, n+2,...")
 
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== Problem ==
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==Problem==
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For positive integers <math>m</math> and <math>n</math> such that <math>m+10<n+1</math>, both the mean and the median of the set <math>\{m, m+4, m+10, n+1, n+2, 2n\}</math> are equal to <math>n</math>. What is <math>m+n</math>?
  
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For positive integers <math>m</math> and <math>n</math> such that <math>m+10<n+1</math>, both the mean and the median of the set <math>\{m, m+4, m+10, n+1, n+2, 2n\}</math> are equal to <math>n</math>. What is <math>m+n</math>?
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<math>\textbf{(A)}20\qquad\textbf{(B)}21\qquad\textbf{(C)}22\qquad\textbf{(D)}23\qquad\textbf{(E)}24</math>
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==Solution 1==
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The mean and median are <cmath>\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,</cmath>so <math>3m+17=2n</math> and <math>m+11=n</math>. Solving this gives <math>\left(m,n\right)=\left(5,16\right)</math> for <math>m+n=\boxed{21}</math>. (trumpeter)
  
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<math>\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24</math>
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==Solution 2==
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This is an alternate solution if you don't want to solve using algebra. First, notice that the median <math>n</math> is the average of <math>m+10</math> and <math>n+1</math>. Therefore, <math>n=m+11</math>, so the answer is <math>m+n=2m+11</math>, which must be odd. This leaves two remaining options: <math>{(B) 21}</math> and <math>{(D) 23}</math>. Notice that if the answer is <math>(B)</math>, then <math>m</math> is odd, while <math>m</math> is even if the answer is <math>(D)</math>. Since the average of the set is an integer <math>n</math>, the sum of the terms must be even. <math>4+10+1+2+2n</math> is odd by definition, so we know that <math>3m+2n</math> must also be odd, thus with a few simple calculations <math>m</math> is odd. Because all other answers have been eliminated, <math>(B)</math> is the only possibility left. Therefore, <math>m+n=\boxed{21}</math>. ∎ --anna0kear
  
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== Solution ==
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==Solution 3==
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Since the median is <math>n</math>, then <math>\frac{m+10+n+1}{2} = n \Rightarrow m+11 = n</math>, or <math>m= n-11</math>. Plug this in for <math>m</math> values to get <math>\frac{7n-16}{6} = n \Rightarrow 7n-16 = 6n \Rightarrow n= 16</math>. Plug it back in to get <math>m = 5</math>, thus <math>16 + 5 = \boxed{21}</math>.
  
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~ iron
  
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The mean and median are <cmath>\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,</cmath>so <math>3m+17=2n</math> and <math>m+11=n</math>. Solving this gives <math>\left(m,n\right)=\left(5,16\right)</math> for <math>m+n=\boxed{21}</math>. (trumpeter)
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== Video Solution 1 ==
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https://youtu.be/5dWRO1-OZOM
  
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~Education, the Study of Everything
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2018|ab=A|num-b=5|num-a=7}}
 
{{AMC12 box|year=2018|ab=A|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:56, 28 October 2022

Problem

For positive integers $m$ and $n$ such that $m+10<n+1$, both the mean and the median of the set $\{m, m+4, m+10, n+1, n+2, 2n\}$ are equal to $n$. What is $m+n$?

$\textbf{(A)}20\qquad\textbf{(B)}21\qquad\textbf{(C)}22\qquad\textbf{(D)}23\qquad\textbf{(E)}24$

Solution 1

The mean and median are \[\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,\]so $3m+17=2n$ and $m+11=n$. Solving this gives $\left(m,n\right)=\left(5,16\right)$ for $m+n=\boxed{21}$. (trumpeter)

Solution 2

This is an alternate solution if you don't want to solve using algebra. First, notice that the median $n$ is the average of $m+10$ and $n+1$. Therefore, $n=m+11$, so the answer is $m+n=2m+11$, which must be odd. This leaves two remaining options: ${(B) 21}$ and ${(D) 23}$. Notice that if the answer is $(B)$, then $m$ is odd, while $m$ is even if the answer is $(D)$. Since the average of the set is an integer $n$, the sum of the terms must be even. $4+10+1+2+2n$ is odd by definition, so we know that $3m+2n$ must also be odd, thus with a few simple calculations $m$ is odd. Because all other answers have been eliminated, $(B)$ is the only possibility left. Therefore, $m+n=\boxed{21}$. ∎ --anna0kear

Solution 3

Since the median is $n$, then $\frac{m+10+n+1}{2} = n \Rightarrow m+11 = n$, or $m= n-11$. Plug this in for $m$ values to get $\frac{7n-16}{6} = n \Rightarrow 7n-16 = 6n \Rightarrow n= 16$. Plug it back in to get $m = 5$, thus $16 + 5 = \boxed{21}$.

~ iron

Video Solution 1

https://youtu.be/5dWRO1-OZOM

~Education, the Study of Everything

See Also

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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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