Find a,b and c. Any method? Given that $$\frac{1}{a+\frac{1}{b+\frac{1}{c+1}}}=\frac{16}{38}$$, find $a,b,c$.
I've been figuring for this quite some time? Is it possible to solve? 
 A: This is a "continued fraction," and there are standard ways of computing them.  To get you started, 
$$\frac{16}{38} = \frac{1}{\frac{38}{16}} = \frac{1}{2+\frac{3}{8}}. $$
So $a=2$.  Continue in this fashion with $\frac{3}{8}$.   Wiki "continued fraction" for (tons) more information.
A: Here is a solution  ; sorry for hand writing 
$$\begin{align*}
\frac{1}{a+\frac{1}{b+\frac{1}{c+1}}}&=\frac{16}{38}\\
a+\frac{1}{b+\frac{1}{c+1}}&=\frac{38}{16}= 2+\frac{3}{8}\\
\text{So}\quad      a=2\\
\frac{1}{b+\frac{1}{c+1}}&=\frac{3}{8}\\
b+\frac{1}{c+1}&=\frac{8}{3}=2+\frac{2}{3}\\
\text{So}\quad     b=2\\
\frac{1}{c+1}&=\frac{2}{3}\\
c+1=\frac{3}{2}=1+\frac{1}{2}\\
\text{So}\quad     c=\frac{1}{2}\\
\end{align*}$$
A: Sure it is, but you'll end up with infinite possibilities:
$$\begin{align*}
\frac{1}{a+\frac{1}{b+\frac{1}{c+1}}}&=\frac{16}{38}\\
a+\frac{1}{b+\frac{1}{c+1}}&=\frac{38}{16}\\
\frac{1}{b+\frac{1}{c+1}}&=\frac{38-16a}{16}\\
b+\frac{1}{c+1}&=\frac{16}{38-16a}\\
\frac{1}{c+1}&=\frac{16-(38-16a)b}{38-16a}\\
c&=\frac{38-16a}{16-(38-16a)b}-1
\end{align*}$$
You can chose $a$ and $b$ freely, and then $c$ will be determined by the two.
