If $a,b,c$ are positive and integers, $a+\frac{1}{b+\frac{1}{c}}=\frac{25}{19}$, then $a+b+c=?$

Question

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If a,b,c are positive and integers, $a+\frac{1}{b+\frac{1}{c}}=\frac{25}{19}$, then $a+b+c=?$ and how to get it?

Answer 1

Hint:

$$a < a+\frac{1}{b+\frac1c} <a+1$$

Then do the same with $$b < b+\frac1c \leq b+1$$

Answer 2

$\begin{eqnarray*} \frac{25}{19} &=& 1+\frac{6}{19} \\[5pt] &=& 1+\frac{1}{\frac{19}{6}} \\[5pt] &=& 1+\frac{1}{3+\frac{1}{6}} \\ \end{eqnarray*}$

$\therefore a+b+c=1+3+6=10$

P.S. The above work shows the existence of $a,b,c$. For the uniqueness see also Ex. 1 on p.3 of http://www.math.hawaii.edu/~pavel/contfrac.pdf