2008 AIME II Problem 8

Question

Let

a = \frac{\pi}{2008}.

Find the smallest positive integer

n

such that

\begin{aligned} &2[\cos(a)\sin(a) + \cos(4a)\sin(2a) \\ &\quad {}+ \cos(9a)\sin(3a) \\ &\quad {}+ \cdots + \cos(n^2 a)\sin(na)] \end{aligned}

is an integer.

Answer

Apply the product-to-sum identity to turn each term into a difference of two sine terms

The sum telescopes to

\sin(n(n+1)a),

which is an integer exactly when

n(n+1)

is a multiple of

1004

By the product-to-sum identity,

\begin{aligned} &2\cos(k^2 a)\sin(ka) \\ &= \sin(k^2 a + ka) \\ &\quad {}- \sin(k^2 a - ka) \\ &= \sin(k(k+1)a) \\ &\quad {}- \sin((k-1)k a). \end{aligned}

Summing over

k = 1

to

n,

the terms telescope, leaving

\sin(n(n+1)a) = \sin\frac{n(n+1)\pi}{2008}.

A sine is an integer only when it is

-1,

0,

or

1,

that is, when its argument is a multiple of

\frac{\pi}{2}.

So we need

\frac{n(n+1)}{2008}

to be a multiple of

\frac{1}{2},

i.e.

n(n+1)

is divisible by

1004,

where

1004 = 4 \cdot 251

and

251

is prime.

Since

n

and

n + 1

are coprime,

251

must divide one of them, so

n \ge 250.

For

n = 250

the product

250 \cdot 251

is not divisible by

4.

For

n = 251

the product

251 \cdot 252

is divisible by

4 \cdot 251 = 1004.

The smallest such

n

is

251.

Problem 8 in other years