<s>[INST] Floor functions in a set of numbers Find the number of distinct numbers in the list
$$\left\lfloor \frac{1^2}{1000} \right\rfloor, \ \left\lfloor \frac{2^2}{1000} \right\rfloor, \ \left\lfloor \frac{3^2}{1000} \right\rfloor, \ \dots, \ \left\lfloor \frac{1000^2}{1000} \right\rfloor.$$
The floor of all numbers till $\left\lfloor\frac{32^2}{1000}\right\rfloor$ are all $0.$ Assuming this goes in cycles of $32,$ there will be $31$ cycles of the $32,$ making there $\boxed{31}$ distinct numbers in this list. Is this the right track?
 [/INST] Consider $\lfloor \frac{x^2}{1000} \rfloor = k$ for some integer $k$. This means:
$$k\leq \frac{x^2}{1000}<k+1$$
$$1000k\leq x^2 \leq 1000k+999$$
It is clear that when $x\geq 500$, $(x+1)^2-x^2=2x+1>1000$ so each value of $\lfloor\frac{x^2}{1000}\rfloor$ is distinct. That's $\fbox{501}$ distinct values from $x=500,501,\cdots,1000$
For $x < 500$ i.e. $k < 250$, all $k$ can be satisfied. $k$ ranges from $0$ to $249$ here, so $\fbox{250}$ distinct values here.
Therefore the answer is $250+501=751$.
</s>