If $\frac{61}{19}=3+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}$ where $x, y$ and $z$ are natural numbers, then what is $z$ equal to?

$\frac{61}{19}=3 \frac{4}{19}=3+\frac{4}{19}$

$\therefore \frac{61}{19}=3+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}$

$\Rightarrow 3+\frac{4}{19}=3+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}$

$\Rightarrow \frac{4}{19}=\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{1}{x+\frac{1}{\frac{y z+1}{z}}}$

$=\frac{1}{x+\frac{z}{y z+1}}=\frac{1}{\frac{x y z+x+z}{y z+1}}$

$\Rightarrow \frac{4}{19}=\frac{y z+1}{x y z+x+z}$

$\therefore y z+1=4 \Rightarrow y z=3$

$x y z+x+z=19$

$\Rightarrow 3 x+x+z=19$

$\Rightarrow 4 x+z=19$

$\Rightarrow x=4, z=3$