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.

Solutions Collecting From Web of "Determine variables that fit this criterion…"

$$
\begin{align}
\frac{25}{84} &= \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc} \\
&= \frac{bc+c+1}{abc} = \frac{(b+1)c+1}{abc}
\end{align}
$$

Now $ 84 = 2\cdot 2 \cdot 3 \cdot 7$ so $b,c$. Let us try $(b+1)c= c = 24 = 2\cdot 2\cdot 2\cdot 3$ and $abc = 84$ (this is just a guess, it can also be a multiple each time). We are looking for $a,b,c$ that are in the factorization of $84$. So $b+1=8$ and $c =3$ would work with $b=7$ since $3$ and $7$ are in the factorization of $84$. So with $abc =84=2\cdot 2 \cdot 3 \cdot 7 = a\cdot 7\cdot 3$ it follows that $a = 4$.

So $\frac{25}{84} = \frac{1}{4} + \frac{1}{28} + \frac{1}{84}$.

Another approach would be trying to find an Eyptian fraction expansion, for example:

$$\frac{25}{84} = \frac{1}{4}+\frac{1}{21} = \frac{1}{4}+\frac{1}{22}+\frac{1}{462} = \frac{1}{4}+\frac{1}{24}+\frac{1}{168} $$

I noticed that in flawr’s answer, b>c. This doesn’t fit the requirement a≤b≤c.

(I don’t have enough reputation to add a comment so I’m posting this as an answer for now)

I think the correct answer is a = 4, b = 6, c = 7.

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 $c=7$.

Clearly $\displaystyle \frac{1}{a} < \frac{25}{84} < \frac{1}{3}$ so $a \geq 4$.
Then $b \geq 4$ and one of $a,b,c$ must be a multiple of $7$ so $c \geq 7$.

Hence $\displaystyle \frac{25}{84} \leq \frac{1}{a} + \frac{1}{4a} + \frac{1}{28a} = \frac{9}{7a}$ so $\displaystyle a \leq \frac{108}{25}<5$ so we must have $a = 4$.

Substitute $a=4$ into the original equation to obtain $\displaystyle \frac{c+1}{bc} = \frac{4}{21}.$ It follows that $4 \mid (c+1)$ and, since $c$ and $c+1$ are coprime, $c \mid 21$. Since we know that $c \geq 7$, we must have that $c = 7$ and so $b = 6$.

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 $(a,b,c)=(4,6,7)$