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Since $p$, $q$, and $r$ must all be positive, $p = 1$ because if it was any higher, $q$ and $r$ would need to be negative to make equation true. So we have $1+\frac{1}{q+\frac{1}{r}}=\frac{25}{19}$. So $\frac{1}{q+\frac{1}{r}}=\frac{6}{19}= \frac{6}{6q+\frac{6}{r}}$, so $19=6q+\frac{6}{r}$. The only possibility of $q$ is $\boxed{3}$, with $r$ being $6$.