Difference between revisions of "2018 AMC 12A Problems/Problem 6"
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| − | == Problem == | + | ==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>? | 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>? | ||
| − | <math>\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24</math> | + | <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== | ||
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) | 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) | ||
| + | ==Solution 2== | ||
| + | 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>. | ||
==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 21:58, 8 February 2018
Contents
Problem
For positive integers 
and 
such that 
, both the mean and the median of the set 
are equal to . What is 
?
Solution 1
The mean and median are ![\[\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,\]](http://latex.artofproblemsolving.com/3/f/f/3ff618fec2b3f9d461b341273cfffab03d55dda9.png)
so 
and 
. Solving this gives 
for 
. (trumpeter)
Solution 2
You can immediately notice that the median is the average of 
and 
. There fore, 
, so now we know we just are looking for 
, 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 
, then is odd, while the opposite is true for if we get 
. 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 is an integer, the sum of the terms must be even. 
is odd by definition, so we know that 
must also be odd, thus with some simple calculations is odd. As with the previous few observations, we have eliminated all other answers, and is the only remaining possibility left. Therefore 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
