A cute problem form India's UPSC exam

 UPSC is India's civil services exam, and surprisingly, the following cute problem appeared. Admittedly, it was a multiple choice question, and could be guessed easily.

Anyway, here is the question:

$a,b,c$ are positive integers such that 

$$\dfrac{bc + 1}{abc + a + c} = \frac{11}{43}$$

Then what are the possible values of $abc$?

Solution.

We can find all solutions. In fact there is a unique solution.

Take the reciprocal and do some minor algebra to get

$$a + \dfrac{c}{bc + 1} = \dfrac{43}{11} = 3 + \dfrac{10}{11}$$ 

Since $\frac{c}{bc + 1} \lt 1$, we must have that $a=3$ and $\frac{c}{bc+1} = \frac{10}{11}$.

Taking reciprocal again gives

$$b + \dfrac{1}{c} = \dfrac{11}{10} = 1 + \dfrac{1}{10}$$

$c = 1$ is not possible (LHS is an integer, while RHS is not). Thus $\frac{1}{c} \lt 1$ and so

$b = 1$ and $c=10$

Thus $(a,b,c) = (3,1,10)$ is the unique solution and $abc = 30$.