Floors and square roots

Let $f(x) = \left \lfloor \sqrt {\left \lfloor \sqrt {\left \lfloor \sqrt {x} \right \rfloor} \right \rfloor} \right \rfloor$ and let us find the the smallest $x$ such that $f(x)=3$ . We note that this occurs when $x$ is an integer and it is equal to $3^8$ such that $\left \lfloor \sqrt {\left \lfloor \sqrt {\left \lfloor \sqrt {3^8} \right \rfloor} \right \rfloor} \right \rfloor = \left \lfloor \sqrt {\left \lfloor \sqrt {3^4} \right \rfloor} \right \rfloor = \left \lfloor \sqrt {3^2} \right \rfloor = 3$ . Since $3^8 = 6561$ is the smallest $x$ such that $f(x) = 3$ , for $x < 6561$ , $f(x) < 3$ . Therefore, the largest integer $x$ such that $f(x)=2$ is $6561-1=\boxed{6560}$ .

This looks difficult by the statement... Let us see how to approach it :

I should not end up getting 3 atlast but 1 number less than that saves me because I will be using floor to get my required answer.

So, if I square 3, I will get 9 and again I will get 81 and again I will get 6561.

So, If I have to get only 2 as my answer I can reduce it only by 1 which is 6560 which gives me 2 because I use floor everytime.