If positive integers $p,q,r$ are such that the quadratic equation $px^2-qx+r=0$ has two distinct real roots in the open interval $(0,1)$, find the minimum value of $p$.
To find the minimum value of \( p \) such that the quadratic equation \( px^2 - qx + r = 0 \) has two distinct real roots in the open interval \( (0,1) \), we need to ensure that the roots satisfy the following conditions:
1. The roots are real and distinct.
2. The roots lie within the interval \( (0,1) \).

Given the quadratic equation \( px^2 - qx + r = 0 \), the roots can be found using the quadratic formula:
\[
x = \frac{q \pm \sqrt{q^2 - 4pr}}{2p}
\]

For the roots to be real and distinct, the discriminant must be positive:
\[
q^2 - 4pr > 0
\]

For the roots to lie within the interval \( (0,1) \), we need:
\[
0 < \frac{q - \sqrt{q^2 - 4pr}}{2p} < 1 \quad \text{and} \quad 0 < \frac{q + \sqrt{q^2 - 4pr}}{2p} < 1
\]

Let's analyze these conditions step by step.

1. **Ensuring the roots are positive:**
\[
\frac{q - \sqrt{q^2 - 4pr}}{2p} > 0 \quad \Rightarrow \quad q - \sqrt{q^2 - 4pr} > 0 \quad \Rightarrow \quad q > \sqrt{q^2 - 4pr}
\]
Since \( q \) and \( \sqrt{q^2 - 4pr} \) are both positive, this inequality is always true.

2. **Ensuring the roots are less than 1:**
\[
\frac{q + \sqrt{q^2 - 4pr}}{2p} < 1 \quad \Rightarrow \quad q + \sqrt{q^2 - 4pr} < 2p
\]
\[
q + \sqrt{q^2 - 4pr} < 2p \quad \Rightarrow \quad \sqrt{q^2 - 4pr} < 2p - q
\]
\[
q^2 - 4pr < (2p - q)^2 \quad \Rightarrow \quad q^2 - 4pr < 4p^2 - 4pq + q^2
\]
\[
-4pr < 4p^2 - 4pq \quad \Rightarrow \quad pr > p^2 - pq
\]
\[
r > p - q
\]

3. **Finding the minimum value of \( p \):**
We need to find the smallest \( p \) such that there exist positive integers \( q \) and \( r \) satisfying the above conditions. Let's test small values of \( p \):

- For \( p = 1 \):
  \[
  r > 1 - q \quad \text{(impossible since \( r \) and \( q \) are positive integers)}
  \]

- For \( p = 2 \):
  \[
  r > 2 - q \quad \text{(impossible since \( r \) and \( q \) are positive integers)}
  \]

- For \( p = 3 \):
  \[
  r > 3 - q \quad \text{(impossible since \( r \) and \( q \) are positive integers)}
  \]

- For \( p = 4 \):
  \[
  r > 4 - q \quad \text{(impossible since \( r \) and \( q \) are positive integers)}
  \]

- For \( p = 5 \):
  \[
  r > 5 - q
  \]
  Let's try \( q = 5 \) and \( r = 1 \):
  \[
  5x^2 - 5x + 1 = 0
  \]
  The discriminant is:
  \[
  5^2 - 4 \cdot 5 \cdot 1 = 25 - 20 = 5 > 0
  \]
  The roots are:
  \[
  x = \frac{5 \pm \sqrt{5}}{10}
  \]
  Both roots are in the interval \( (0,1) \).

Thus, the minimum value of \( p \) is \( 5 \).

The final answer is \( \boxed{5} \).