9. Let $p$ and $q$ be integers such that $p+q=2010$. If both roots of the equation $10 x^{2}+p x+q=0$ are positive integers, find the sum of all possible values of $p$.
(1 mark)
設 $p 、 q$ 為整數, 且 $p+q=2010$ 。若方程 $10 x^{2}+p x+q=0$ 的兩根均為正整數, 求 $p$ 所有可能值之和。
（1 分）
9. -3100
9. Let $\alpha$ and $\beta$ be the roots of the equation $10 x^{2}+p x+q=0$. Then we have $\alpha+\beta=-\frac{p}{10}$ and $\alpha \beta=\frac{q}{10}$, and so $2010=p+q=(-10)(\alpha+\beta)+10 \alpha \beta$, which gives $\alpha \beta-\alpha-\beta=201$ or $(\alpha-1)(\beta-1)=202$. As $\alpha$ and $\beta$ are positive integers, the two factors $\alpha-1$ and $\beta-1$ may be 1 and 202 , or 2 and 101 . Hence $\alpha$ and $\beta$ may be 2 and 203, or 3 and 102. These correspond to $p=-2050$ and $p=-1050$ respectively. It follows that the answer is -3100 .