1. If $a, b$ are both integers, the equation
$$
a x^{2}+b x-2008=0
$$

has two distinct roots that are prime numbers, then $3 a+b=$ $\qquad$ (2008, Taiyuan Junior High School Mathematics Competition)
Let the two prime roots of the equation be \(x_{1} 、 x_{2}\left(x_{1}<x_{2}\right)\). From the problem, we have
\[
x_{1} x_{2}=\frac{-2008}{a} \Rightarrow a x_{1} x_{2}=-2008 \text{. }
\]

It is easy to see that, \(2008=2^{3} \times 251\) (251 is a prime number).
Thus, \(x_{1}=2, x_{2}=251\).
Therefore, \(3 a+b=1000\).