6. If $a, b$ are integers, and the equation $a x^{2}+b x-$ $2008=0$ has two distinct prime roots, then the value of $3 a+b$ is ( ).
(A) 100
(B) 400
(C) 700
(D) 1000
6.D.

Let the roots of the equation be $x_{1}, x_{2}\left(x_{1}<x_{2}\right)$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a}, \\
x_{1} x_{2}=-\frac{2008}{a},
\end{array}\right.
$$

which means
$$
\left\{\begin{array}{l}
b=-a\left(x_{1}+x_{2}\right), \\
a x_{1} x_{2}=-2008=-2 \times 2 \times 2 \times 251.
\end{array}\right.
$$
Since $x_{1}, x_{2}$ are prime numbers, we can take $x_{1}=2$, $x_{2}=251$, from which we get $a=-4$.
And $b=4(2+251)=1012$, thus
$$
3 a+b=1000.
$$