Difference between revisions of "2014 AMC 12A Problems/Problem 21"

Latest revision as of 14:38, 12 June 2024

Problem

For every real number $x$ , let $\lfloor x\rfloor$ denote the greatest integer not exceeding $x$ , and let $f(x)=\lfloor x\rfloor(2014^{x-\lfloor x\rfloor}-1).$ The set of all numbers $x$ such that $1\leq x<2014$ and $f(x)\leq 1$ is a union of disjoint intervals. What is the sum of the lengths of those intervals?

$\textbf{(A) }1\qquad \textbf{(B) }\dfrac{\log 2015}{\log 2014}\qquad \textbf{(C) }\dfrac{\log 2014}{\log 2013}\qquad \textbf{(D) }\dfrac{2014}{2013}\qquad \textbf{(E) }2014^{\frac1{2014}}\qquad$

Solution

Let $\lfloor x\rfloor=k$ for some integer $1\leq k\leq 2013$ . Then we can rewrite $f(x)$ as $k(2014^{x-k}-1)$ . In order for this to be less than or equal to $1$ , we need $2014^{x-k}-1\leq\dfrac1k\implies x\leq k+\log_{2014}\left(\dfrac{k+1}k\right)$ . Combining this with the fact that $\lfloor x\rfloor =k$ gives that $x\in\left[k,k+\log_{2014}\left(\dfrac{k+1}k\right)\right]$ , and so the length of the interval is $\log_{2014}\left(\dfrac{k+1}k\right)$ . 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 $k=1$ to $k=2013$ to get that the desired sum is $\sum_{i=1}^{2013}\log_{2014}\left(\dfrac{i+1}i\right)=\log_{2014}\left(\prod_{i=1}^{2013}\dfrac{i+1}i\right)=\log_{2014}\left(\dfrac{2014}1\right)=\boxed{1\textbf{ (A)}}.$

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2014amc12a/380

~ dolphin7

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=MI3ax4WJBZA

(The video is no longer available on YouTube)

@@ Line 19: / Line 19: @@
 ==Video Solution by Punxsutawney Phil==
 https://youtube.com/watch?v=MI3ax4WJBZA
+(The video is no longer available on YouTube)
 ==See Also==
 {{AMC12 box|year=2014|ab=A|num-b=20|num-a=22}}
 {{MAA Notice}}
+[[Category:Intermediate Algebra Problems]]

2014 AMC 12A (Problems • Answer Key • Resources)
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