Determine variables that fit this criterion... There is a unique triplet of positive integers $(a, b, c)$ such that $a ≤ b ≤ c$.
$$
\frac{25}{84} = \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc}
$$
Just having trouble with this Canadian Math Olympiad question. My thought process going into this, is:
Could we solve for $\frac{1}{a}$ in terms of the other variables? Then substitute that value in for each occurrence of $a$, to solve for $a$?
That's all I can really think of right now. It's a question I'm not exactly used to... It's sort of the first of these kinds that I've faced.
Thanks.

Factoring, we see
$\displaystyle \frac{25}{84} = \frac{1}{a}(1+\frac{1}{b}(1+\frac{1}{c}))$
And we know the prime factoring of 84 gives $2\times2\times3\times7$  So we know $a,b,$ and $c$ are each going to be multiples of these primes.  So we start with finding $a$:
$\displaystyle \frac{25}{84}a = 1+\frac{1}{b}(1+\frac{1}{c})$
Now, $25a/84>0$, but $\frac{1}{b}(1+\frac{1}{c})>0$ too.  Therefore $25a/84>1$.  We want $a$ to be the smallest of the three factors, so we ask, what is the smallest it can be here?  2 won't work and neither will 3, but $2\times2=4$ will.  So we provisionally say $a=4$.  Then,
$\displaystyle \frac{25}{21} = 1+\frac{1}{b}(1+\frac{1}{c})$
We go through the same process for $b$, remembering that $b>4$.  Turns out that $b=3\times2=6$ is the smallest factor that will work.  We provisionally say $b=6$.
Finally,
$\displaystyle \frac{8}{7} = 1+\frac{1}{c}$
$c=7$ follows immediately.
So $a=4$, $b=6$, and $c=7$.
