There is a polynomial $P(x)$ with integer coefficients such that\[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\]holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$.
Because $0 < x < 1$, we have
\begin{align*} P \left( x \right) & = \sum_{a=0}^6  \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \binom{6}{a} x^{2310a} \left( - 1 \right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\ & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} x^{2310 a + 105 b + 70 c + 42 d + 30 e} . \end{align*}
Denote by $c_{2022}$ the coefficient of $P \left( x \right)$.
Thus,
\begin{align*} c_{2022} & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} \Bbb I \left\{ 2310 a + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & =  \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-0} \Bbb I \left\{ 2310 \cdot 0 + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \Bbb I \left\{ 105 b + 70 c + 42 d + 30 e = 2022 \right\} . \end{align*}
Now, we need to find the number of nonnegative integer tuples $\left( b , c , d , e \right)$ that satisfy 
\[ 105 b + 70 c + 42 d + 30 e = 2022 . \hspace{1cm} (1) \]
Modulo 2 on Equation (1), we have $b \equiv 0 \pmod{2}$.
Hence, we can write $b = 2 b'$. Plugging this into (1), the problem reduces to finding the number of 
nonnegative integer tuples $\left( b' , c , d , e \right)$ that satisfy
\[ 105 b' + 35 c + 21 d + 15 e = 1011 . \hspace{1cm} (2) \]
Modulo 3 on Equation (2), we have $2 c \equiv 0 \pmod{3}$.
Hence, we can write $c = 3 c'$. Plugging this into (2), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d , e \right)$ that satisfy
\[ 35 b' + 35 c' + 7 d + 5 e = 337 . \hspace{1cm} (3) \]
Modulo 5 on Equation (3), we have $2 d \equiv 2 \pmod{5}$.
Hence, we can write $d = 5 d' + 1$. Plugging this into (3), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e \right)$ that satisfy
\[ 7 b' + 7 c' + 7 d' + e = 66 . \hspace{1cm} (4) \]
Modulo 7 on Equation (4), we have $e \equiv 3 \pmod{7}$.
Hence, we can write $e = 7 e' + 3$. Plugging this into (4), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e' \right)$ that satisfy
\[ b' + c' + d' + e' = 9 . \hspace{1cm} (5) \]
The number of nonnegative integer solutions to Equation (5) is $\binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} =  \boxed{\textbf{(220) }}$.
~Steven Chen (www.professorchenedu.com)