2016 AMC 12B Problems/Problem 2

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 $\boxed{(A)}$

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 12 Problems and Solutions

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2016 AMC 12B Problems/Problem 2

Problem

Solution

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