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Let $z,w$ be complex and equal to $e^{2\pi i/101}$ and $ e^{2\pi i/10}$, respectively.

Show that $$\prod_{a=0}^9 \prod_{b=0}^{100}\prod_{c=0}^{100}(w^a+z^b+z^c)$$ is an integer and find the remainder upon division by $101$.

I tried converting the product to a sum by logarithms but I really couldn't go any further.

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    how did you come about this problem? – 2017-01-12
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    Long time ago in a math competition far far away. – 2017-01-13
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    So I played with this for a bit, and I have an idea. Consider resultants, and then think about how to create a product with 3 terms instead of 2. I think the answer is 0 or 1 btw (as a gut instinct) – 2017-01-22
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    Mathematica says 13, but I couldn't care less about the answer. I need to know how to get there. – 2017-01-23
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    This seems like a very tricky question to ask on a competition. – 2017-01-23
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    I think this will be answered soon, but I did manage to reduce the number to $$ \prod_{b, c}^{100} \[(z^b + z^c)^{10} - 1 \] $$ – 2017-02-03
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    I gave this to my friend @Oliver Clarke, and he says he has an idea and will probably solve it by tomo, so we'll see how that goes – 2017-02-03

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