How to prove $\prod _{a=0}^{9}\prod_{b=0}^{100}\prod_{c=0}^{100}(w^a+z^b+z^c)\equiv?\pmod {101}$ Let $z=e^{\dfrac{2\pi i}{101}},w=e^{\dfrac{2\pi i}{10}}$,show that
$$A=\prod _{a=0}^{9}\prod_{b=0}^{100}\prod_{c=0}^{100}(w^a+z^b+z^c)$$
 find $A\equiv ?\pmod {101}$
I guess this problem maybe is USA problem,but I can't find it,and I can't solve it(this problem is from people sent to me)
 A: No official sources, but this agrees with iF9n's (half-numerical) solution here. If $\omega$ is a primitive $n$th root of unity, then $$\prod_{k=0}^{n-1} (x - \omega^k) = x^n - 1$$ and so $$\prod_{k=0}^{n-1} (x + \omega^k) = (-1)^n \prod_{k=0}^{n-1} (-x - \omega^k) = (-1)^n ((-x)^n - 1) = x^n + (-1)^{n+1}.$$
Now we have $$\begin{align*} \prod_{a=0}^9 \prod_{b=0}^{100} \prod_{c=0}^{100} (w^a + z^b + z^c) &= \prod_{a=0}^9 \prod_{b=0}^{100} ( (w^a + z^b)^{101} + 1) \\ &\equiv \prod_{a=0}^9 \prod_{b=0}^{100} (w^{101a} + z^{101b} + 1) \pmod {101} \\
&= \prod_{a=0}^9 \prod_{b=0}^{100} (w^{a} + 2) \\
&= \left[ \prod_{a=0}^9 (w^a + 2) \right]^{101} \end{align*}$$
because $w^{100} = z^{101} = 1$ and the binomial coefficients $p \choose n$ for $p$ prime, besides ${p \choose 0} = {p \choose p} = 1$, are multiples of $p$. 
The modular reduction here is subtle, as the binomial expansion of $(w^a + z^b)^{101}$ contains non-integer terms. But every term in the expansion, except $w^{101a}$ and $z^{101b}$, is $101$ times some element in $\mathbb{Z}[w, z] = \mathbb{Z}[\omega_{1010}]$, where $\omega_n$ is an $n$th primitive root of unity. Thus, we can interpret "reduction modulo $101$" as reduction from $\mathbb{Z}[\omega_{1010}]$ to $\mathbb{Z}[\omega_{1010}]/(101)$, which does kill all the middle terms in the binomial expansion. Such reduction may turn a non-integer (like, say, $5 + 101 z$) into an integer, but that's not a problem here: $A$ is an algebraic integer and it's invariant under the Galois automorphisms of $\mathbb{Q}[\omega_{1010}]/\mathbb{Q}$ given by $\omega_{1010} \mapsto \omega_{1010}^k$ (which also take $\omega_{10} \mapsto \omega_{10}^k$ and $\omega_{101} \mapsto \omega_{101}^k$ where $\gcd(k, 1010) = 1$—thanks to Erick Wong in comments for pointing this out), so $A \in \mathbb{Q} \cap \mathbb{Z}[\omega_{1010}] = \mathbb{Z}$, and reduction modulo $101$ won't spuriously turn $A$ into an integer.
Finally, $$\prod_{a=0}^9 (2+w^a) = 2^{10} - 1 = 1023,$$ so the whole expression (by Fermat's little theorem) is congruent to $1023^{101} \equiv 1023 \equiv 13 \pmod {101}$.
