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.

NOTE:When doing Olympiad problems you're given a lot of time,so writing answers like this might not be as time consuming as it would be on a normal exam.This answer is big but after noticing $c$ is a divisor of 84 rest is easy,though I'll try to find a simpler,quicker solution by tomorrow
$$\frac{25}{84}=\frac{c(b+1)+1}{abc}$$
Now from this we can conclude that $gcd(c,c(b+1)+1)=1$ so $c$ has to be a divisor of $84$,if $c<7$ than at least one of b,c has to be a multiple of $7$ and hence bigger than $c$ so if we plug $c=7$ if $b=6$ than $a=4$,if $b$ is smaller by inspection it doesn't fit.$c=12$ than either $a$ or $b$ are equal to $7$ if $b=7$ than numerator is prime if $a=7$ than $25=c+\frac{c+1}{b}$ so $b=1$ but $a>b$ now if $c=84$ than $$25ab=84b+85\\b(25a-84)=85$$ from this we get that $b=5k$ since $25a-84$ isn't divisible by 5 from this $k(25a-84)=17$ if $a>4$ left side is bigger than the right side so no solutions now if $c=42$ we have that $$25ab=84b+86\\b(25a-84)=86=2*43$$
Since $b\not=43,86$ we can say $b=1,2$ than $a<4$ which is not possible from now on we assume $b\geq4$ case $c=28$
$$25ab=84b+87\\b(25a-84)=87=3*29$$
Either $a,b$ are multiples of $3$ and both $a,b$ are odd,$a=3$ and $a=9$ so $a$ isn't divisible by $3$ and $b=3$ doesn't fit
If $c=21$ than $$25ab=84b+88\\b(25a-84)=88=2^3*11$$
$b=4$ is not solution and for $b=8,11$ $25a-84\geq16$ while RHS$<16$
Now if $c=14$
$$25ab=84b+90\\b(25a-84)=90$$
Now $b=5k$ since $25a-84$ isn't divisible by 5,$b=5,10$ do not fit so the only solution is $(a,b,c)=(4,6,7)$
