Recursive problem

The answer is 10. First, it is easy to see that $a=1$ , since $\frac{25}{19}=1+\frac{6}{19}$ . Now, we have $\frac{1}{b+\frac{1}{c}} = \frac{6}{19}$ . If we invert the whole equation, we have $b + \frac{1}{c} = \frac{19}{6} = 3 + \frac{1}{6}$ . Using the same logic to find $a$ , we have that $b=3$ and $c=6$ . The answer follows.

We know that a , b and c are all positive integers.

$\frac{1}{b + \frac{1}{c}} = \frac{1}{\frac{bc + 1}{c}} = \frac{c}{bc + 1}$

$a + \frac{c}{bc + 1} = \frac{abc + a + c}{bc + 1}$

We know that $\frac{abc + a + c}{bc + 1}$ is equal to $\frac{25}{19}$ , so we firstly need to find b and c that solve bc + 1 = 19.

bc + 1 = 19 then bc = 18.

There are four different combinations of integers now:

1. b = 2 and c = 9

2. b = 9 and c = 2

3. b = 6 and c = 3

4. b = 3 and c = 6

You can try the other combinations, but the right one is b = 3 and c = 6.

Knowing this, in our first equation:

$\frac{abc + a + c}{bc + 1} = \frac{25}{19}$

18a + + 6 = 25

18a + a = 19

From which we can tell a = 1.

Finally, a + b + c = 1 + 3 + 6 = 10.