<s>[INST] Hints to find solution of : $\left\lfloor \frac{x}{100} \left\lfloor \frac{x}{100}\right\rfloor \right\rfloor=5$ Find the number of integer solution of $$\left\lfloor  \dfrac{x}{100} \left\lfloor \dfrac{x}{100}\right\rfloor \right\rfloor=5$$
$\lfloor . \rfloor$ is Greatest Integer Function or floor function.
 [/INST] 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 $x$.
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