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2008 AIME II Problems/Problem 8

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Problem

Let a = \pi/2008. Find the smallest positive integer n such that 2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)] is an integer.

Solution

By the product-to-sum identities, we have that 2\cos a \sin b = \sin (a+b) - \sin (a-b). Therefore, this reduces to a telescoping series: \begin{align*}\sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} [\sin(k(k+1)a) - \sin((k-1)ka)]\\&= -\sin(0) + \si...

Thus, we need \sin \left(\frac{n(n+1)\pi}{2008}\right) to be an integer; this can be only \{-1,0,1\}, which occur when 2 \cdot \frac{n(n+1)}{2008} is an integer. Thus 1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1. It easily follows that n = \boxed{251} is the smallest such integer.

See also

2008 AIME II (Problems • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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