1. **State the problem:** Calculate the value of the summation $$\sum_{n=3}^{2017} \sin \left( \frac{(n!)\pi}{36} \right)$$. 2. **Analyze the sine argument:** The summation term is $$\sin \left( \frac{(n!)\pi}{36} \right)$$ where $n!$ is factorial of $n$. Since factorial grows very rapidly, consider the sine function periodicity. 3. **Periodicity of sine:** Sine has period $2\pi$, so: $$\sin(\theta) = \sin(\theta + 2k\pi) \quad \text{for any integer } k.$$ 4. **Simplify factorial modulo the period denominator:** We want to find $$\sin \left( \frac{(n!)\pi}{36} \right) = \sin \left( \pi \times \frac{n!}{36} \right).$$ Since sine is periodic with period $2\pi$, the argument can be reduced modulo $2\pi$: $$\frac{(n!)\pi}{36} \equiv \frac{(n!)\pi}{36} \mod 2\pi.$$ Dividing both sides by $\pi$, this is equivalent to reducing $\frac{n!}{36}$ modulo 2: $$\frac{n!}{36} \equiv x \mod 2 \implies n! \equiv 36x \mod 72.$$ So we want to find $n! \mod 72$. 5. **Check factorial modulo 72:** Since $72 = 8 \times 9$, check when $n!$ is divisible by 72. - For $n=6$, $6! = 720$ which is divisible by 72 since $720/72=10$. - For all $n \geq 6$, $n!$ is divisible by 72 because factorial includes factors 8 and 9 and more. 6. **Calculate sine values for $n=3,4,5$ directly:** - $3! = 6 \, \Rightarrow \, \sin\left( \frac{6\pi}{36} \right) = \sin\left( \frac{\pi}{6} \right) = \frac{1}{2}$ - $4! = 24 \Rightarrow \sin\left( \frac{24\pi}{36} \right) = \sin\left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}$ - $5! = 120 \Rightarrow \sin\left( \frac{120\pi}{36} \right) = \sin\left( \frac{10\pi}{3} \right)$ Reduce $\frac{10\pi}{3}$ modulo $2\pi$: $$\frac{10\pi}{3} - 3\times 2\pi = \frac{10\pi}{3} - 6\pi = \frac{10\pi - 18\pi}{3} = -\frac{8\pi}{3} = 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}.$$ So, $$\sin\left( \frac{10\pi}{3} \right) = \sin\left( \frac{4\pi}{3} \right) = -\frac{\sqrt{3}}{2}.$$ 7. **For $n \geq 6$, since $n!$ divisible by 72:** $$\frac{(n!)\pi}{36} = \frac{72k\pi}{36} = 2k\pi, \quad \text{for some integer } k.$$ Therefore, $$\sin\left( 2k\pi \right) = 0.$$ 8. **Sum the terms:** $$\sum_{n=3}^{2017} \sin \left( \frac{(n!)\pi}{36} \right) = \sin\left( \frac{6\pi}{36} \right) + \sin\left( \frac{24\pi}{36} \right) + \sin\left( \frac{120\pi}{36} \right) + \sum_{n=6}^{2017} 0$$ Which simplifies to: $$\frac{1}{2} + \frac{\sqrt{3}}{2} + \left(-\frac{\sqrt{3}}{2}\right) = \frac{1}{2}.$$ **Final answer:** $$\sum_{n=3}^{2017} \sin \left( \frac{(n!)\pi}{36} \right) = \frac{1}{2}.$$