Problem:
Let a=Ο/2008. Find the smallest positive integer n such that
2[cos(a)sin(a)+cos(4a)sin(2a)+cos(9a)sin(3a)+β―+cos(n2a)sin(na)]
is an integer.
Solution:
The product-to-sum formula for 2cos(k2a)sin(ka) yields sin(k2a+ka)β sin(k2aβka), which equals sin(k(k+1)a)βsin((kβ1)ka). Thus the given sum becomes
sin(2β
1a)ββsin(0)+sin(3β
2a)βsin(2β
1a)+sin(4β
3a)βsin(3β
2a)+β―+sin(n(n+1)a)βsin((nβ1)na).β
This is a telescoping sum, which simplifies to sin(n(n+1)a)βsin(0)= sin(n(n+1)a). But sin(x) is an integer only when x is an integer multiple of Ο/2, so n(n+1) must be an integer multiple of 1004=22β
251. Thus either n or n+1 is a multiple of 251, because 251 is prime. The smallest such n is 250, but 250β
251 is not a multiple of 1004. The next smallest such n is 251, and 251β
252 is a multiple of 1004. Hence the smallest such n is 251β.
The problems on this page are the property of the MAA's American Mathematics Competitions