CEMC Fermat, 2013

Since $p$ is a positive integer, then $p \geq 1$ and so $\text{0 lt; 1 p 1}$.

Since $n$ is a positive integer, then $n \geq 1$ and so $\text{n + 1 p gt;1}$, which tells us that $\text{0 lt; 1 n+ 1 p lt; 1}$.

Therefore, $\text{m lt; m + 1 n+ 1 p lt; m+1}$. 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 $\text{n lt; n+ 1 p n+1}$ 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, $n=1$.