Problem:
For each integer n>1, let F(n) be the number of solutions of the equation sinx=sinnx on the interval [0,Ο]. What is βn=22007βF(n)?
Answer Choices:
A. 2,014,524
B. 2,015,028
C. 2,015,033
D. 2,016,532
E. 2,017,033
Solution:
Note that F(n) is the number of points at which the graphs of y=sinx and y=sinnx intersect on [0,Ο]. For each n,sinnxβ₯0 on each interval [(2kβ2)Ο/n,(2kβ1)Ο/n] where k is a positive integer and 2kβ1β€n. The number of such intervals is n/2 if n is even and (n+1)/2 if n is odd. The graphs intersect twice on each interval unless sinx=1=sinnx at some point in the interval, in which case the graphs intersect once. This last equation is satisfied if and only if nβ‘1(mod4) and the interval contains Ο/2. If n is even, this count does not include the point of intersection at (Ο,0). Therefore F(n)=2(n/2)+1=n+1 if n is even, F(n)=2(n+1)/2=n+1 if nβ‘3 (mod4), and F(n)=n if nβ‘1(mod4). Hence
n=2β2007βF(n)=(n=2β2007β(n+1))ββ42007β1ββ=2(2006)(3+2008)ββ501=2,016,532β.
The problems on this page are the property of the MAA's American Mathematics Competitions