Product of "complex" distances

This problem may be solved by simple "Complex bashing". Notice that all the unitary circumference with its regular polygon in it, may rapresent "roots of unity", or namely the solutions to $x^n=1$ visualized on the Gauss' plane. $D(n)$ corresponds to $|1-x||1-x^2|...|1-x^{n-1}|$ , which indeed turns out to be exactly $n$ . $M(n)$ is always $2$ . An easy way to see this is to recognize that $M(n)=\frac{D(2n)}{D(n)}$ . The sum corresponds now to $\sum\limits_{n=3}^{2021} 2^n \cdot n$ whose $3$ last digits turn out to be $072$ (obtained by differentiating).

Observe that $M(n)=2$ and $D(n)=n$ . So we just want to evaluate $\sum_{n=3}^{2021}2^n \cdot n$

Working (mod 10), observe that the powers of 2 have a periodicity of 4, and n has periodicity of 10, so the last digit of subsequent terms has periodicity LCM(4,10)=20. Work out the first 20 terms: $\sum_{n=3}^{22}2^n\cdot n \equiv 4+4+0+4+6+8+8+0+8+2+6+6+0+6+4+2+2+0+2+8 \equiv 0 \pmod {10}$ Notice that our sum is just lacking one term to complete 101 batches of 20, the answer is $-8 \equiv \boxed{2} \pmod {10}$ .