The number $2017$ is prime.  Let $S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}$.  What is the remainder when $S$ is divided by $2017?$
$\textbf{(A) }32\qquad \textbf{(B) }684\qquad \textbf{(C) }1024\qquad \textbf{(D) }1576\qquad \textbf{(E) }2016\qquad$
Note that $2014\equiv -3 \mod2017$.  We have for $k\ge1$
\[\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017\]
\[\equiv (-1)^k\dbinom{k+2}{k} \mod 2017\]
Therefore 
\[\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017\]
This is simply an alternating series of triangular numbers that goes like this: $1-3+6-10+15-21....$
After finding the first few sums of the series, it becomes apparent that 
\[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv -\left(\frac{n+1}{2} \right) \left(\frac{n+1}{2}+1 \right) \mod 2017 \textnormal{  if n is odd}\]
and 
\[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv \left(\frac{n}{2}+1 \right)^2 \mod 2017 \textnormal{  if n is even}\]
Obviously, $62$ falls in the second category, so our desired value is 
\[\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{\textbf{(C)}\ 1024}\]

Sidenote
Another way to finish, using the fact that $\dbinom{k+2}{2} = 1 + 2 + \dots + (k+1)$:
\begin{align*} \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} &\equiv \sum \limits_{k=1}^{63}(-1)^{k-1} (1 + 2 + \dots + k) \\ &\equiv 1 - (1+2) + (1+2+3) - (1+2+3+4) + \dots + (1 + \dots + 63) \\ &\equiv 1 + 3 + 5 + \dots + 63 \\ &\equiv \boxed{1024} \mod 2017 \end{align*}