, 2013

Since $p$ is a positive integer, then $p \geq 1$ and so $0<\dfrac{1}{p}\leq 1$.

Since $n$ is a positive integer, then $n \geq 1$ and so $n + \dfrac{1}{p}>1$, which tells us that $0 < \dfrac{1}{n+\dfrac{1}{p}} < 1$.

Therefore, $m < m + \dfrac{1}{n+\dfrac{1}{p}} < m+1$. Since $m + \dfrac{1}{n+\dfrac{1}{p}} = \dfrac{17}{3}$, which is between 5 and 6, and since $m$ is an integer, then $m=5$.

Since $m=5$, then $m + \dfrac{1}{n+\dfrac{1}{p}} = \dfrac{17}{3}$ gives $\dfrac{1}{n+\dfrac{1}{p}} = \dfrac{2}{3}$ or $n+\dfrac{1}{p} = \dfrac{3}{2}$.

Since $n < n+\dfrac{1}{p} \leq n+1$ and $n$ is an integer, then $n=1$.

Thus, $n+\dfrac{1}{p} = \dfrac{3}{2}$ gives $\dfrac{1}{p} = \dfrac{1}{2}$, which gives $p=2$.

Therefore, $n=1$.