6. (1995 National High School Mathematics Competition) 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.


6. Solution: Clearly, $x=1$ is a natural number root of the original equation. By synthetic division, we get $5 x^{2}-5 p x+66 p-1=0$.

Thus, the original proposition is equivalent to (1) having two natural number roots, denoted as $\mu, \gamma(\mu \leqslant \gamma)$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
\mu+\gamma=p, \\
\mu \gamma=\frac{1}{5}(66 p-1) .
\end{array}\right.
$$
(2) Substituting (2) into (3) yields $5 \mu \gamma=66(\mu+\gamma)-1$.

From (4), we know that $\mu, \gamma$ cannot be divisible by $2, 3, 11$. Also from (4), we get $\gamma=\frac{66 \mu-1}{5 \mu-66}$ (5). Since $\mu, \gamma$ are natural numbers, from (5) we know $\mu>\frac{66}{5}$, i.e., $\mu \geqslant 14$. Furthermore, since $2 \times \mu$ and $3 \times \mu$, we have $\mu \geqslant 17$.

Given $\gamma \geqslant 4$ and (5), we have $\frac{66 \mu-1}{5 \mu-66} \geqslant \mu$. Thus, $5 \mu^{2}-132 \mu+1 \leqslant 0$. Therefore, $\mu \leqslant \frac{66+\sqrt{66^{2}-5}}{5}<\frac{132}{5}$, so $17 \leqslant \mu \leqslant 26$. Since $2 \times \mu$ and $3 \times \mu$, $\mu$ can only be $17, 19, 23, 25$. After testing, only $\mu=17, \gamma=59$ satisfy the conditions, hence $p=76$.