$12 \cdot 144$ Find all real numbers $p$ such that the cubic equation
$$
5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p
$$

has three roots that are all natural numbers.
(China High School Mathematics League, 1995)
[Solution 1] Since $5-5(p+1)+(77 p-1)+1=66 p$, then $x=1$ is a natural number solution to the original cubic equation.
By synthetic division, the original cubic equation is reduced to a quadratic equation
$$
5 x^{2}-5 p x+66 p-1=0
$$

This problem is converted to: find all real numbers $p$, such that equation (1) has two natural number solutions.
Let $u, v(u \leqslant v)$ be the two natural number solutions of equation (1). By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
v+u=p \\
v u=\frac{1}{5}(66 p-1) .
\end{array}\right.
$$

Eliminating $p$ from (2) and (3) yields
$$
5 u v=66(u+v)-1 .
$$

From (4), it is known that $u, v$ cannot be divisible by $2,3,11$.
From (4), we get $v=\frac{66 u-1}{5 u-66}$.
Since $u, v$ are natural numbers, then
$$
u>\frac{66}{5} \text {, so } u \geqslant 14 \text {. }
$$

Also, $2 \times u, 3 \times u$, so $u \geqslant 17$.
Furthermore, by $v \geqslant u$, we get $\quad \frac{66 u-1}{5 u-66} \geqslant u$,

which means
$$
5 u^{2}-132 u+1 \leqslant 0 \text {. }
$$

Thus, $u \leqslant \frac{66+\sqrt{66^{2}-5}}{5}_{q}$, leading to $\left\{\begin{array}{l}m_{0}-q_{0}=1,19,229,4351, \\ m_{0}+q_{0}=4351,229,19,1 .\end{array}\right.$

Solving this, we get
$$
m_{0}= \pm 2176, \pm 124 \text {. }
$$

Therefore, $5 p-132=2 m_{0}= \pm 4352, \pm 248$.
From (2) in Solution 1, we know that $p$ is a natural number, and from (3), we know that $u, v$ are odd numbers, and by (2), we know that $p$ is even.
Thus, we can solve to get $p=76$.