Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime) positive integers. Find $m + n$.

note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques
We can rewrite $xyz=1$ as $\frac{1}{z}=xy$.
Substituting into one of the given equations, we have 
\[x+xy=5\]
\[x(1+y)=5\]
\[\frac{1}{x}=\frac{1+y}{5}.\]
We can substitute back into $y+\frac{1}{x}=29$ to obtain
\[y+\frac{1+y}{5}=29\]
\[5y+1+y=145\]
\[y=24.\]
We can then substitute once again to get
\[x=\frac15\]
\[z=\frac{5}{24}.\]
Thus, $z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$, so $m+n=\boxed{005}$.