Problem
2020 Fermat Problem 2
The value of 3 \times 2020 + 2 \times 2020 - 4 \times 2020 is
\textbf{(A)}\ 6060\quad \textbf{(B)}\ 4040\quad \textbf{(C)}\ 8080\quad \textbf{(D)}\ 0\quad \textbf{(E)}\ 2020
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