23. Given that the number 2017 is a prime number, let $S=\sum_{k=0}^{62} \mathrm{C}_{2014}^{k}$. Then the remainder when $S$ is divided by 2017 is ().
(A) 32
(B) 684
(C) 1024
(D) 1576
(E) 2016
23. C.

Notice,
$$
\begin{array}{l}
\mathrm{C}_{2014}^{k}=\frac{2014 \times 2013 \times \cdots \times(2014-k+1)}{k!} \\
\equiv \frac{(-3) \times(-4) \times \cdots \times(-2-k)}{k!} \\
\equiv(-1)^{k} \mathrm{C}_{2+k}^{k} \equiv(-1)^{k} \mathrm{C}_{2+k}^{2}(\bmod 2017) . \\
\text { Therefore, } S=\sum_{k=0}^{62} \mathrm{C}_{2014}^{k} \equiv \sum_{k=0}^{62}(-1)^{k} \mathrm{C}_{2+k}^{2} \\
\equiv \mathrm{C}_{2}^{2}-\mathrm{C}_{3}^{2}+\mathrm{C}_{4}^{2}-\mathrm{C}_{5}^{2}+\cdots-\mathrm{C}_{63}^{2}+\mathrm{C}_{64}^{2} \\
\equiv \frac{2 \times 1}{2}-\frac{3 \times 2}{2}+\frac{4 \times 3}{2}-\frac{5 \times 4}{2}+\cdots-\frac{63 \times 62}{2}+\frac{64 \times 63}{2} \\
\equiv \frac{2 \times 1}{2}+\left(-\frac{3 \times 2}{2}+\frac{4 \times 3}{2}\right)+ \\
\left(-\frac{5 \times 4}{2}+\frac{6 \times 5}{2}\right)+\cdots+\left(-\frac{63 \times 62}{2}+\frac{64 \times 63}{2}\right) \\
\equiv 1+3+5+\cdots+63 \equiv 32^{2} \\
\equiv 1024(\bmod 2017) .
\end{array}
$$