For any real number $t$, let $[t]$ (or $\lfloor t\rfloor$) denote the largest integer less than or equal to $t$. Suppose that $N$ is the greatest integer such that $\lfloor \sqrt{\lfloor \sqrt{\lfloor \sqrt{N}\rfloor }\rfloor }\rfloor =4$. Find the sum of the digits of $N$.

Let the nested floor equation be:

We want to find the greatest integer $N$ satisfying this.
Let $x=\lfloor \sqrt{\lfloor \sqrt{N}\rfloor }\rfloor$. The equation is $\lfloor \sqrt{x}\rfloor =4$.
This implies:

Since $x$ is an integer, the maximum value of $x$ is $24$.

Now, $x=\lfloor \sqrt{\lfloor \sqrt{N}\rfloor }\rfloor =24$. This implies:

So the maximum value of $\lfloor \sqrt{N}\rfloor$ is $624$.

Finally, $\lfloor \sqrt{N}\rfloor =624$. This implies:

The greatest integer $N$ is $625^2-1=390625-1=390624$.

The sum of the digits of $N=390624$ is: