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Geometry Level 3

$\large \displaystyle \sum^{2017}_{n=3} \sin \left( \dfrac{(n!)\pi}{36} \right)$

Find the value of the above sum.

0 1

-\frac{1}{2}

\frac{1}{2}

1 solution

Achal Jain
Jun 20, 2020

We can use the fact that for all $n>=6$ , $n!$ $mod$ $36=0$ . Thus we only need to compute from $n=3$ to $n=5$ .