Find the number of sets of $(a,b,c)$ for $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{29}{72}$ 
If $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{29}{72},\ \ c<b<a<60,\ \ \{a,b,c\}\in\mathbb{N} $.
How many sets of $(a,b,c)$ exists ?

Options
$a.)\ 3 \quad \quad \quad \quad \quad b.)\ 4 \\
c.)\ 5 \quad \quad \quad \quad \quad \color{green}{d.)\ 6} \\ $ 
by trial and error i found 
$$\begin{array}{c|c} 
2 & 72 \\ \hline
2 &36  \\ \hline
2 &18 \\ \hline
3 &9 \\ \hline
3 &3 \\ \hline
 &1\\
 \end{array}$$
$\\~\\~\\$
$$\dfrac{29}{72}=\dfrac{18}{72}+\dfrac{9}{72}+\dfrac{2}{72}= \dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a}\\~\\
\implies (a,b,c)=(36,8,4)$$
This question is from chapter quadratic equations.
I look for a short and simple way.
I have studied maths up to $12$th grade.

In this type of problem,
you have to go through cases,
usually on the extreme variables.
Initially,
$\dfrac{1}{c} < \dfrac{29}{72}$,
so
$c > \dfrac{72}{29}
=2+\dfrac{14}{29}
$,
so $c \ge 3$.
In the other direction,
since
$\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a}
< \dfrac{3}{c}
$,
$\dfrac{3}{c} > \dfrac{29}{72}$
or
$c < \dfrac{216}{29}
$
or $c \le 7$.
Looking at $a$,
$\dfrac{3}{a} < \dfrac{29}{72}$,
or $a \ge 8$.
For each $3 \le c \le 7$,
compute
$d
=\dfrac{72}{29}-\dfrac{1}{c}
$.
Then
$\dfrac{1}{a}+\dfrac{1}{b} = d$,
so
$\dfrac{2}{a} < d < \dfrac{2}{b}$
or
$c+1 \le b < \dfrac{2}{d}$.
For each $c+1 \le b < \dfrac{2}{d}$,
compute
$\dfrac{1}{d}-\dfrac{1}{b}$
and see if that
is of the form
$\dfrac{1}{a}$
for $a > b$.
I'll leave the actual computation
to you.
