17. The positive integers $m, n$ and $p$ satisfy the equation $3 m+\frac{3}{n+\frac{1}{p}}=17$. What is the value of $p$ ?
A 2
B 3
C 4
D 6
E 9
Solution
A

Because $\frac{3}{n+\frac{1}{p}}=17-3 m$, where $m$ is an integer, it follows that $\frac{3}{n+\frac{1}{p}}$ is an integer.
Because $n$ and $p$ are positive integers, $1<n+\frac{1}{p}$ and therefore $0<\frac{3}{n+\frac{1}{p}}<3$. Therefore $\frac{3}{n+\frac{1}{p}}$ is either 1 or 2 .
If $\frac{3}{n+\frac{1}{p}}=1$, then $3 m=17-\frac{3}{n+\frac{1}{p}}=17-1=16$. This implies that $m=\frac{16}{3}$ contradicting the fact that $m$ is an integer. Hence $\frac{3}{n+\frac{1}{p}}=2$. Therefore $n+\frac{1}{p}=\frac{3}{2}=1+\frac{1}{2}$. Hence $n=1$ and $p=2$. Also $3 m=17-\frac{3}{1+\frac{1}{2}}=17-2=15$ and so $m=5$. We therefore see that $m=5, n=1$ and $p=2$.