What a Trigonometric Product!

$\sin\Bigl(\frac{\pi}{n}\Bigr)\cos\Bigl(\frac{\pi}{n}\Bigr)\cos\Bigl(\frac{2\pi}{n}\Bigr)\cos\Bigl(\frac{4\pi}{n}\Bigr)\cos\Bigl(\frac{8\pi}{n}\Bigr)\cos\Bigl(\frac{16\pi}{n}\Bigr)$ $=\dfrac{1}{32}\sin \Bigl(\frac{32\pi}{n}\Bigr)$

Therefore the equation can be written as :

$\sin (\dfrac{32\pi}{n})=\sin(\dfrac{\pi}{n})$

or $\cos(\dfrac{33\pi}{2n}) \sin(\dfrac{31\pi}{2n})=0$

So $\dfrac{33\pi}{2n} = \dfrac{\pi}{2} + \pi k$ or $\dfrac{31\pi}{2n} = \pi k$ . The first case yields $n=1,3,11,33$ and the second case yields no solutions. $n=1$ is clearly extraneous, so the answer is $47$