Hints to find solution of : $\left\lfloor \frac{x}{100} \left\lfloor \frac{x}{100}\right\rfloor \right\rfloor=5$

Hint: focusing first on the inner floor, clearly we must have $0 \lt \lfloor \frac x{100} \rfloor \lt 3$, so it can only be $1$ or $2$. If it is $1$, how large can the product inside the outer floor be? If it is $2$, what range of $x$ works?

As $y-1<\lfloor y\rfloor\le y $, we find that for $y\ge1$ $$\tag1y^2-y-1<\lfloor y\lfloor y\rfloor\rfloor\le y^2 $$ whereas for $y<0$, we have $$\tag2y^2-y-1>\lfloor y\lfloor y\rfloor\rfloor\ge y^2 $$ Finally, for $0\le y<1$, we clearly have $\lfloor y\lfloor y\rfloor \rfloor=0$ Thus we cannot have $\lfloor y\lfloor y\rfloor\rfloor=5$ with $y\le -\sqrt 5$ (right hand side of $(2)$), nor with $-2\le y<0$ (left hand side of $(2)$), nor with $0\le y<1$, nor with $1\le y<\sqrt 5$ (right hand side of $(1)$), nor with $y\ge 3$ (left hand side of $(1)$). In other words, $-\sqrt 5< y <-2$ or $\sqrt 5\le y<3$. Then $\lfloor y\rfloor =-3$ or $\lfloor y\rfloor=2$.

In the first case, $$-3y-1<\lfloor -3y\rfloor=\lfloor y\lfloor y\rfloor \rfloor \le -3y,$$ so that we need $y\ge -\frac 53$, which contardicts $y<-2$.

In the second case $$2y-1<\lfloor 2y\rfloor=\lfloor y\lfloor y\rfloor \rfloor \le 2y,$$ so that we need $y\ge\frac 52$. And indeed, for $\frac25\le y<3$, we verify directly that $\lfloor y\lfloor y \rfloor\rfloor=5 $. If we impose that $y=\frac x{100}$ with $x\in\Bbb Z$, this is equivalent to $$250\le x<300$$ whic allows exactly $50$ differnet values of $x$.

First note that as $[n] \le n < [n]+1$

then $[n]^2 \le [n[n]] < ([n]+1)^2$

So $[x/100]^2 \le [n[n]]= 5 < ([x/100] + 1)^2$. As $[x/100]$ is an integer, this means $[x/100] = 2$

So

$[x/100[x/100]]= [x/100*2] = [x/50] = 5$

$5 \le x/50 < 6$

$250 \le x < 300$ so $x = [250.... 299]$.