Problem:
For every real number x, let ⌊x⌋ denote the greatest integer not exceeding x, and let
f(x)=⌊x⌋(2014x−⌊x⌋−1).
The set of all numbers x such that 1≤x<2014 and f(x)≤1 is a union of disjoint intervals. What is the sum of the lengths of those intervals?
Answer Choices:
A. 1
B. log2014log2015
C. log2013log2014
D. 20132014
E. 201420141
Solution:
If x=n+r, where n is an integer, 1≤n≤2013, and 0≤r<1, then f(x)=n(2014r−1). The condition f(x)≤1 is equivalent to 2014r≤1+n1, or 0≤r≤log2014(nn+1). Thus the required sum is
log201412+log201423+log201434+⋯+log201420132014=log2014(12⋅23⋅34⋯20132014)=log2014(2014)=1
The problems on this page are the property of the MAA's American Mathematics Competitions