23. Find the largest real number $p$ such that all three roots of the equation below are positive integers:
$$
5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p .
$$
23. Answer: 76
Solution. Observe that $x=1$ is always a root of the equation
$$
5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p .
$$

Thus this equation has all roots positive integers if and only if the two roots of the equation below are positive integers:
$$
5 x^{2}-5 p x+66 p-1=0 .
$$

Let $u, v$ be the two roots with $u \leq v$. Then
$$
u+v=p_{4} \quad u v=(66 p-1) / 5,
$$
implying that
$$
5 u v=66(u+v)-1 .
$$

By this expression, we know that $u, v$ are not divisible by any one of $2,3,11$. We also have $5 u v>66(u+v)$, implying that
$$
\frac{2}{u} \geq \frac{1}{u}+\frac{1}{v}>\frac{5}{66}
$$
and so $u \leq 26$. As
$$
v=\frac{66 u-1}{5 u-66}>0,
$$
we have $5 u-66>0$ and so $u \geq 14$. Since $u$ is not a multiple of any one of $2,3,11$, we have
$$
u \in\{17,19,23,25\} \text {. }
$$

As $v=\frac{60 \alpha-1}{5 \alpha-60}$, only when $u=17, v=59$ is an integer.
Thus, only when $p=u+v=17+69=76$, the equation
$$
5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p
$$
has all three roots being positive integers.