35. Let $p$ be an integer such that both roots of the equation
$$
5 x^{2}-5 p x+(66 p-1)=0
$$
are positive integers. Find the value of $p$.
35. Ans: 76
Let $u, v$ be the two positive integral solutions of the given equation. Then
$$
\left\{\begin{array}{l}
u+v=p, \\
u v=\frac{66 p-1}{5} .
\end{array}\right.
$$

Upon eliminating $p$, we have
$$
\begin{aligned}
5 u v=66(u+v)-1 & \Longrightarrow v(5 u-66)=66 u-1>0 \\
& \Longrightarrow 5 u-66>0, \text { since } v>0 .
\end{aligned}
$$

Similarly, we have $5 v-66>0$. Moreover, we have
$$
\begin{aligned}
(5 u-66)(5 v-66) & =25 u v-330(u+v)+66^{2} \\
& =25 \times\left(\frac{66 p-1}{5}\right)-330 p+66^{2} \\
& =-5+66^{2}=4351=19 \times 229 .
\end{aligned}
$$

Without loss of generality, we may assume that $u \geq v$. Since both 19 and 229 are prime, we must have
$$
\left\{\begin{array} { l l } 
{ 5 u - 6 6 = 2 2 9 } \\
{ 5 v - 6 6 = 1 9 }
\end{array} \text { or } \left\{\begin{array}{ll}
5 u-66=4351 \\
5 v-16 & =1 .
\end{array}\right.\right.
$$

The first set of equations imply $u=59, v=17$. The second set of equations does not have integral solutions. Hence, we must have $u=59$ and $v=17$. Thus, $p=u+v=76$.