# Difference between revisions of "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 $\boxed{(A)}$