$X$ is an odd number, $Y$ is a natural number more than 36. If $\frac{1}{X}+\frac{2}{Y}=\frac{1}{18}$, find the set $(X,Y)$? $X$ is an odd number, $Y$ is a natural number more than 36. If $\frac{1}{X}+\frac{2}{Y}=\frac{1}{18}$, find the set $(X,Y)$ ?
Re arranging the given equation, we have,
$\frac{2}{Y}=\frac{X-18}{18*X}$
$\frac{1}{Y}=\frac{X-18}{36*X}$
$Y=\frac{36*X}{X-18}$
$Y=\frac{1*2*2*3*3*X}{X-18}$
Now we know that $X$ is an odd number, therefore $X-18$ must also be an odd number.
By visual inspection we see that, $X-18$ can take values of 1, 3 and 9
i.e. $X-18 = 1 = 3^{0}$
     and $X=19$
$X-18 = 3 = 3^{1}$
i.e, $X=21$
$X-18 = 9 = 3^{2}$
i.e, $X=27$
when I see the pattern for $X-18$ to be $3^{0}, 3^{1}, 3^{2}$ I'm tempted to try for $3^{3}, 3^{4},...$
when $X-18=3^3$ we have $X=45$
when $X-18=3^4$ we have $X=99$
I get the answer but whats the reasoning behind this ?
Is there another way to solve this question ?

As you have $Y=\frac{36X}{X-18},$
So, $Y=\frac{36(X-18)+36\cdot 18}{X-18}=36+\frac{3^4\cdot2^3}{X-18}$
also as, $Y>0, X>18$ 
and as you have identified $X-18=3^r ,0\le r\le 4 $

Alternatively
HINT:
So, $Y=\frac{36X}{X-18}$ which is even $=2Z$(say)
So, we have $$\frac1 X+\frac1 Z=\frac1{18}\implies XZ-18(X+Z)=0$$
$$\implies (X-18)(Z-18)=18^2\implies Z-18=\frac{18^2}{X-18}\text{ which is an integer }$$
Clearly, $(Z-18)(X-18)\ne0\implies X,Z>18$
