$2(81+83+85+87+89+91+93+95+97+99)=$
$\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800$
Find that
\[(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180\]
Which gives us
\begin{align*} 2(5 \cdot 180) &= 10 \cdot 180\\ &= 1800 & \text{ Thus \boxed{\text{E}} is the correct answer} \end{align*}