Difference between revisions of "2016 AMC 12B Problems/Problem 2"

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which is finally closest to <math>\boxed{\textbf{(A)}\ 2}</math>.
 
which is finally closest to <math>\boxed{\textbf{(A)}\ 2}</math>.
 
 
-dragonfly
 
-dragonfly
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+
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You can also think of <math>\frac{2\times1\times2016}{1+2016}</math> as <math>2\times\frac{2016}{2017}</math> which it should be obvious that our number is most closest to 2 (<math>\lceil\frac{2016}{2017}\rceil=1</math>), our answer here is <math>\boxed{(A)}</math>
 
You can also think of <math>\frac{2\times1\times2016}{1+2016}</math> as <math>2\times\frac{2016}{2017}</math> which it should be obvious that our number is most closest to 2 (<math>\lceil\frac{2016}{2017}\rceil=1</math>), our answer here is <math>\boxed{(A)}</math>
  

Latest revision as of 16:47, 25 January 2021

Problem

The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of $1$ and $2016$ is closest to which integer?

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 504 \qquad \textbf{(D)}\ 1008 \qquad \textbf{(E)}\ 2015$

Solution

Since the harmonic mean is $2$ times their product divided by their sum, we get the equation

$\frac{2\times1\times2016}{1+2016}$

which is then

$\frac{4032}{2017}$

which is finally closest to $\boxed{\textbf{(A)}\ 2}$. -dragonfly


You can also think of $\frac{2\times1\times2016}{1+2016}$ as $2\times\frac{2016}{2017}$ which it should be obvious that our number is most closest to 2 ($\lceil\frac{2016}{2017}\rceil=1$), our answer here is $\boxed{(A)}$

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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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