Difference between revisions of "2014 AMC 12A Problems/Problem 21"
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==Solution== | ==Solution== | ||
Let <math>\lfloor x\rfloor=k</math> for some integer <math>1\leq k\leq 2013</math>. Then we can rewrite <math>f(x)</math> as <math>k(2014^{x-k}-1)</math>. In order for this to be less than or equal to <math>1</math>, we need <math>2014^{x-k}-1\leq\dfrac1k\implies x\leq k+\log_{2014}\left(\dfrac{k+1}k\right)</math>. Combining this with the fact that <math>\lfloor x\rfloor =k</math> gives that <math>x\in\left[k,k+\log_{2014}\left(\dfrac{k+1}k\right)\right]</math>, and so the length of the interval is <math>\log_{2014}\left(\dfrac{k+1}k\right)</math>. 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 <math>k=1</math> to <math>k=2013</math> to get that the desired sum is <cmath>\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)}}.</cmath> | Let <math>\lfloor x\rfloor=k</math> for some integer <math>1\leq k\leq 2013</math>. Then we can rewrite <math>f(x)</math> as <math>k(2014^{x-k}-1)</math>. In order for this to be less than or equal to <math>1</math>, we need <math>2014^{x-k}-1\leq\dfrac1k\implies x\leq k+\log_{2014}\left(\dfrac{k+1}k\right)</math>. Combining this with the fact that <math>\lfloor x\rfloor =k</math> gives that <math>x\in\left[k,k+\log_{2014}\left(\dfrac{k+1}k\right)\right]</math>, and so the length of the interval is <math>\log_{2014}\left(\dfrac{k+1}k\right)</math>. 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 <math>k=1</math> to <math>k=2013</math> to get that the desired sum is <cmath>\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)}}.</cmath> | ||
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+ | ==See Also== | ||
+ | {{AMC12 box|year=2014|ab=A|num-b=20|num-a=22}} | ||
+ | {{MAA Notice}} |
Revision as of 19:00, 7 February 2014
Problem
For every real number , let
denote the greatest integer not exceeding
, and let
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
, 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.