Weird question pertaining to HCF I encountered this question which seems weird/incomplete to me :

Q: H.C.F. of 3240, 3600 and a third number is 36, and their L.C.M. is
  $2^4 \cdot 3^5 \cdot 5^2 \cdot 7^2$ . The third number is?

Can anyone please teach me concept wise how to solve it? 

\begin{align}
3240 & = 2\cdot2\cdot2\cdot3\cdot3\cdot3\cdot3\cdot5 & & = 2^3\cdot3^4\cdot5\\
3600 & = 2\cdot2\cdot2\cdot2\cdot3\cdot3\cdot5\cdot5 & & = 2^4\cdot3^2\cdot5^2 \\
36 & = 2\cdot2\cdot3\cdot3 & & =2^2\cdot3^2
\end{align}
The third number cannot have more than two $2$s in its prime factorization, since then the gcd would not be $36$, but at least $2$ times that.  The third cannot have any $5$ in its prime factorization, since then the gcd would have a $5$ in it.  The third number must have two $7$s in its prime factorization, since there's nothing else to contribute those two $7$s to the lcm.  The third number must have at least two $2$s in its prime factoriation; otherwise it would not be divisible by $36$.  The third number must have at least two $3$s in its prime factorization for the same reason.
So the third number could be $2\cdot2\cdot3\cdot3\cdot7\cdot7$.
It could also be $2\cdot2\cdot3\cdot3\cdot3\cdot7\cdot7$ or $2\cdot2\cdot3\cdot3\cdot3\cdot3\cdot7\cdot7$.
