2014 AMC 12A Problems/Problem 21
Contents
Problem
For every real number 
, let 
denote the greatest integer not exceeding , and let ![\[f(x)=\lfloor x\rfloor(2014^{x-\lfloor x\rfloor}-1).\]](http://latex.artofproblemsolving.com/b/6/d/b6d01f35e21f4dda60a2836b577dcbd218144d36.png)
The set of all numbers such that 
and 
is a union of disjoint intervals. What is the sum of the lengths of those intervals?
Solution
Let 
for some integer 
. Then we can rewrite 
as 
. In order for this to be less than or equal to 
, we need 
. Combining this with the fact that 
gives that ![$x\in\left[k,k+\log_{2014}\left(\dfrac{k+1}k\right)\right]$](http://latex.artofproblemsolving.com/f/1/6/f166eefdf28e0d71694031656161e036fb398b86.png)
, and so the length of the interval is 
. We want the sum of all possible intervals such that the inequality holds true; since all of these intervals must be disjoint, we can sum from 
to 
to get that the desired sum is ![\[\sum_{i=1}^{2013}\log_{2014}\left(\dfrac{i+1}i\right)=\log_{2014}\left(\prod_{i=1}^{2013}\dfrac{i+1}i\right)=\log_{2014}\left(\dfrac{2014}1\right)=\boxed{1\textbf{ (A)}}.\]](http://latex.artofproblemsolving.com/0/6/2/06261d18b68a17761cb48e3ae0a8834f53752c6b.png)
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2014amc12a/380
~ dolphin7
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=MI3ax4WJBZA
See Also
| 2014 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
Followed by Problem 22 |
| 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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