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

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==Solution 2==
 
==Solution 2==
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You can immediately notice that the median <math>n</math> is the average of <math>m+10</math> and <math>n+1</math>. There fore, <math>n=m+11</math>, so now we know we just are looking for <math>m+n=2m+11</math>, which must be odd. This leaves our two remaining options, {(B)}21\qquad\textbf and {(D)}23\qquad\textbf. Note that if the answer is <math>(B)</math>, then <math>m</math> is odd, while the opposite is true for <math>m</math> if we get <math>(D)</math>. This fact will come in handy later on and prove itself useful when we solve for its parity. Since the average of the set of six numbers <math>n</math> is an integer, 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 some simple calculations <math>m</math> is odd. As with the previous few observations, we have eliminated all other answers, and <math>(B)</math> is the only remaining possibility left. Therefore <math>m+n=(B)\boxed{21}</math>.
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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
  
 
==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 23:07, 8 February 2018

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

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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