The sum of all real numbers $$p$$ such that the equation $$5x^3 - 5(p+1)x^2 + (71p-1)x - (66p-1) = 0$$ has all its three roots positive integers.

If the roots are positive integers $$r, s, t$$ then $$r+s+t = p+1$$, $$rs+st+tr = \frac{71p-1}{5}$$ and $$rst = \frac{66p-1}{5}$$. Eliminating $$p$$ and searching the resulting integer conditions gives the single possibility $$r, s, t = 1, 17, 59$$ with $$p = 76$$, since $$1+17+59 = 77 = p+1$$, $$1079 = \frac{71 \times 76 - 1}{5}$$ and $$1003 = \frac{66 \times 76 - 1}{5}$$. As this is the only such value, the required sum is 76.