1. (Shanghai $\cdot$ Senior High School) If positive integers $p, q, r$ make the quadratic equation $p x^{2}-{ }_{y} x+r=0$ have two distinct real roots in the open interval $(0,1)$, find the minimum value of $p$.

Let $\alpha, \beta$ be the two roots of the quadratic equation $p x^{2}-q x+\gamma=0$, $\alpha \neq \beta, 0<\alpha, \beta<1$. Then
$$
\alpha \beta(1-\alpha)(1-\beta) = \frac{r}{p} \left[1 - \frac{q}{p} + \frac{r}{p}\right] = \frac{r}{p^{2}}(p-q+r).
$$

Therefore, $p^{2} > 16 r(p-q+r)$.
Since the quadratic term coefficient $p > 0$ of $f(x) = p x^{2} - q x + r$, and its roots are within $(0,1)$, we have $f(1) = (p - q + r) = 0$, and $r(p - q + r) > 0$. Given that $p, q, r$ are positive integers, we have $r(p - q + r) \geq 1$. Thus, from the inequality, we get
$$
p^{2} > 16 r(p - q + r) = 16.
$$

Therefore, $p > 1$, i.e., $p > 5$.
When $p = 5$, the quadratic equation $5 x^{2} - 5 x + 1 = 0$ has two roots $\alpha = \frac{5 + \sqrt{5}}{10}, \beta = \frac{5 - \sqrt{5}}{10}$, both of which are within $(0,1)$. Hence, $p_{10} = 5$.