A typical complex number problem 1

dont actually deserves level 4 its level 5!!

firstly we would derive the product of first n terms product of {sin(kpi/n)} k is from 1 to n-1

we use complex here we know that the nth root of unity are of the e^{2pik/n) and are 1,a1,--------a(n-1)

using property of nth root of unity(you may derive the property by transformation of eqn)

|(1-a1)(1-a2)(1-a2)-------(1-a(n-1))|=n

|1-ak|=|1-{cos(2kpi/n)+i*sin(2kpi/n)}|=2sin(kpi/n)

2^(n-1){ product of {sin(kpi/n)} k is from 1 to n-1}=n

*product of {sin(kpi/n)} k is from 1 to n-1= $\frac{n}{2^{n-1}}$ *

but gcd(k,n)=1

so our q is $\frac{[product of {sin(kpi/100)} k is from 1 to 99]}{[sin(2pi/100)sin(4pi/100)sin(5pi/100)----]}$

= $\frac{[product of {sin(kpi/100)} k is from 1 to 99]*[sin(10pi/100)sin(20pi/100)---}{[{sin(2pi/100)sin(4pi/100)----}{sin(5pi/100)sin(25pi/100)----}]}$

(since we counted those k which are a multiple of 10 (2 times) in 2 and 5 series.)

== $\frac{[product of {sin(kpi/100)} k is from 1 to 99]*[[product of {sin(kpi/10)} k is from 1 to 9]}{[{[product of {sin(kpi/50)} k is from 1 to 49]}[product of {sin(kpi/20)} k is from 1 to 19]]}$

now applying the general formula we derived we get ans = 2^(-40)

One way to do this is to use the formula $\prod_{k=1}^n \sin(k\pi/n) = \frac{n}{2^{n-1}}.$ Then we can use Mobius inversion . We have $\prod_{d|n}P_d = \frac{n}{2^{n-1}}$ so $\prod_{d|n} \left(\frac{d}{2^{d-1}}\right)^{\mu(n/d)} = P_n.$ For $n=100$ this is easy to compute as $1/2^{40},$ so the answer is $\fbox{-40}.$

In general I think we can get pretty far to get to the general solution. First, $P_n = \frac{\prod d^{\mu(n/d)}}{2^{\sum (d-1)\mu(n/d)}},$ and the denominator is easy to compute as $2^{\phi(n)}$ (unless $n=1,$ in which case it's $1$ ).

The numerator is $g(n)$ where $\prod g(d) = n,$ and it's easy to see that in that case $g(d) = e^{\Lambda(d)}$ where $\Lambda(n)$ is von Mangoldt's function; i.e. $g(n) = p$ if $n$ is a power of $p$ and $1$ otherwise.

So our final answer, for $n > 1,$ is $P_n = \frac{e^{\Lambda(n)}}{2^{\phi(n)}} = \begin{cases} \frac{p}{2^{\phi(n)}} &\text{ if } n =p^k, p \text{ prime} \\ \frac1{2^{\phi(n)}} &\text{ otherwise.} \end{cases}$ I think this is right but I might have missed something.