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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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 
and 
is closest to which integer?
Solution
Since the harmonic mean is 
times their product divided by their sum, we get the equation
which is then
which is finally closest to 
.
-dragonfly
You can also think of as 
which it should be obvious that our number is most closest to 2 (
), our answer here is 
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
