5. Given $f(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+a_{3} x^{2}+a_{4} x+a_{5}$, and when $m=1,2,3,4$, $f(m)=2017 m$, then $f(10)-f(-5)=$
A. 71655
B. 75156
C. 75615
D. 76515
Given that $1,2,3,4, a$ are the 5 roots of $f(x)-2017 x$, then
$$
\begin{array}{l}
f(x)=(x-1)(x-2)(x-3)(x-4)(x-a)+2017 x, \text { thus } \\
f(10)-f(-5)=[6 \cdot 7 \cdot 8 \cdot 9(10-a)+2017 \cdot 10]-[6 \cdot 7 \cdot 8 \cdot 9(-5-a)+2017 \cdot(-5)] \\
=(6 \cdot 7 \cdot 8 \cdot 9+2017) \cdot 15=75615, \text { so the answer is } C .
\end{array}
$$