Difference between revisions of "2018 AMC 12A Problems/Problem 6"
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− | == Problem == | + | ==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, 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> | |
+ | |||
+ | ==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{\textbf{(B)}~21}</math>. (trumpeter) | ||
− | <math> | + | ==Solution 2== |
+ | 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 | ||
− | == Solution == | + | ==Solution 3== |
+ | 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{\text{(B)}~21}</math>. | ||
+ | ~ iron | ||
− | + | == Video Solution 1 == | |
+ | https://youtu.be/5dWRO1-OZOM | ||
+ | ~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}} |
Latest revision as of 17:32, 5 August 2024
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 so
and
. Solving this gives
for
. (trumpeter)
Solution 2
This is an alternate solution if you don't want to solve using algebra. First, notice that the median is the average of
and
. Therefore,
, so the answer is
, which must be odd. This leaves two remaining options:
and
. Notice that if the answer is
, then
is odd, while
is even if the answer is
. Since the average of the set 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 a few simple calculations
is odd. Because all other answers have been eliminated,
is the only possibility left. Therefore,
. ∎ --anna0kear
Solution 3
Since the median is , then
, or
. Plug this in for
values to get
. Plug it back in to get
, thus
.
~ iron
Video Solution 1
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.