How many positive values of $a$ are possible in $2^{3}\le a\lfloor a\rfloor \le 4^{2} + 1$ How many positive values of $a$ are possible in the following case?
$$2^{3}\le a\lfloor a\rfloor \le 4^{2} + 1$$
where $a\lfloor a\rfloor$ such that $a[a]$ is an integer.

It is clear that $\lfloor a\rfloor$ must be $\ge 3$ and $\le 4$.
In the case $\lfloor a\rfloor=3$, the product can be $9$, $10$, or $11$.  For $8$ is too small, since $a\ge \lfloor a\rfloor$. And $12$ is too big, since then $a=4$, giving the wrong value for the floor function. So $a$ can have values $3$, $10/3$, and $11/3$.
In the case $\lfloor a\rfloor=4$ there are two possible values of the product, $16$ and $17$.
So the number of possible values of $a$ is $5$.
