2014 AMC 12A Problem 21

Question

For every real number

x,

let

\lfloor x\rfloor

denote the greatest integer not exceeding

x,

and let

f(x)=\lfloor x\rfloor\left(2014^{\,x-\lfloor x\rfloor}-1\right).

The set of all numbers

x

such that

1\le x\lt2014

and

f(x)\le1

is a union of disjoint intervals. What is the sum of the lengths of those intervals?

1

\dfrac{\log2015}{\log2014}

\dfrac{\log2014}{\log2013}

\dfrac{2014}{2013}

2014^{\frac{1}{2014}}

Write

x=n+r

with

n=\lfloor x\rfloor

and

0\le r\lt1,

so

f(x)=n\left(2014^{\,r}-1\right)

The condition

f(x)\le1

gives

0\le r\le\log_{2014}\dfrac{n+1}{n}

; sum these lengths and telescope

Write

x=n+r

with integer

n

(

1\le n\le2013

) and

0\le r\lt1.

Then

f(x)=n\left(2014^{\,r}-1\right),

and

f(x)\le1

becomes

2014^{\,r}\le1+\dfrac1n,

i.e.

0\le r\le\log_{2014}\dfrac{n+1}{n}.

Each

n

contributes an interval of length

\log_{2014}\dfrac{n+1}{n},

so the total is

\begin{gathered} \sum_{n=1}^{2013}\log_{2014}\dfrac{n+1}{n}\\ =\log_{2014}\!\left(\dfrac21\cdot\dfrac32\cdots\dfrac{2014}{2013}\right)\\ =\log_{2014}2014=1. \end{gathered}

Thus, the correct answer is A.

Problem 21 in other years