Find the number of different numbers of the form $\left\lfloor\frac{i^2}{2015} \right\rfloor$, with $i = 1,2, ..., 2015$.
Problem
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Tags: floor function, combinatorics, number theory, algebra
fuzimiao2013
05.10.2021 21:42
Realize that $\lfloor \frac{1^2}{2015} \rfloor = 0$ and $\lfloor \frac{2015^2}{2015} \rfloor = 2015$, and so all numbers from $0$ to $2015$, inclusive, work. Thus, the number of different numbers of that form is 2016.
pinkpig
05.10.2021 21:53
parmenides51 wrote: Find the number of different numbers of the form $\left\lfloor \frac{i^2}{2015}\right\rfloor$, with $i = 1,2, ..., 2015$. $\Big \lfloor \Big \rfloor$ does not work. You have to use $\left\lfloor\right\rfloor$
peace09
05.10.2021 22:02
I understand this part: Quote: Realize that $\lfloor \frac{1^2}{2015} \rfloor = 0$ and $\lfloor \frac{2015^2}{2015} \rfloor = 2015$ ... But I don't see how from here, we can conclude the following: Quote: ... and so all numbers from $0$ to $2015$, inclusive, work. Could you elaborate?
fuzimiao2013
05.10.2021 22:40
peace09 wrote: I understand this part: Quote: Realize that $\lfloor \frac{1^2}{2015} \rfloor = 0$ and $\lfloor \frac{2015^2}{2015} \rfloor = 2015$ ... But I don't see how from here, we can conclude the following: Quote: ... and so all numbers from $0$ to $2015$, inclusive, work. Could you elaborate? oh wait. thats wrong, dangit