Difference between revisions of "2018 AMC 12A Problems/Problem 6"
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==Solution 1== | ==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{\textbf{(B)}~21}</math>. (trumpeter) |
==Solution 2== | ==Solution 2== |
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 |
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