Problem:
Let f(x)=∑k=210(⌊kx⌋−k⌊x⌋), where ⌊r⌋ denotes the greatest integer less than or equal to r. How many distinct values does f(x) assume for x≥0?
Answer Choices:
A. 32
B. 36
C. 45
D. 46
E. Infinitely many
Solution:
Note that for any x,f(x+1)=∑k=210(⌊kx+k⌋−k⌊x+1⌋)= ∑k=210(⌊kx⌋+k−k⌊x⌋−k)=f(x). This implies that f(x) is periodic with period 1. Thus the number of distinct values that f(x) assumes is the same as the number of distinct values that f(x) assumes for 0≤x<1. For these x,⌊x⌋=0, so f(x)=∑k=210⌊kx⌋, which is a nondecreasing function of x. This function increases at exactly those values of x expressible as a fraction of positive integers with denominator between 2 and 10. There are 31 such values between 0 and 1. They are 21,31,32,41,43,51,52,53,54,61,65,71,72,73,74,75,76,81,83,85,87,91,92,94,95,97,98,101, 103,107,109. Thus f(0)=0 and f(x) increases 31 times for x between 0 and 1, showing that f(x) assumes (A)32 distinct values.
The problems on this page are the property of the MAA's American Mathematics Competitions