Evaluate $\sum_{r=1}^{502} \lfloor \frac{305r}{503} \rfloor$.

Question

The question is to find the value of —

$$\sum_{r=1}^{502} \Big \lfloor \frac{305r}{503}\Big \rfloor$$

The answer is pretty big, so I don't think trial and error will work here. I seriously can't come up with a solution to this problem. I have asked plenty of floor function questions (you can check my profile), but this one is quiet different. Can someone please help me out? Even a hint is greatly appreciated.

Answer 1

Here is one way: use Pick's theorem on the triangle with vertices $(0,0), (503,0), (503,305)$.

Or you can do it more algebraically by noting $$ \left\lfloor\frac{305 r}{503}\right\rfloor + \left\lfloor\frac{305(503-r)}{503}\right\rfloor = 305-1=304 $$ for all $1\leq r\leq 502$ (since $503$ is prime).

Answer 2

One can prove the following formula (for, say, positive integers $a, b, c$): $$\sum_{k=0}^{b-1}\Big\lfloor\frac{ak+c}{b}\Big\rfloor=\frac{(a-1)(b-1)+d-1}{2}+d\Big\lfloor\frac{c}{d}\Big\rfloor,\qquad d=\gcd(a,b)$$

Answer 3

This is a junior problem (China /1986) check the solution of (example 7,page 41) here, Since all real numbers $x$ and $y$ are not integers but $x+y$ is integer then : $\left\lfloor \sqrt{x}\right\rfloor+\left\lfloor \sqrt{y}\right\rfloor=x+y-1$ since $\{x\}+ \{y\}=1$, Since $\frac{305r}{503}+\frac{305(503-r)}{503}=305, 1\leq r\leq 502$ then we have : \begin{align*} & S=\left\lfloor \frac{305}{503}\right\rfloor+\left\lfloor \frac{305\times 2}{503}\right\rfloor+\cdots +\left\lfloor \frac{305\times 502}{503}\right\rfloor \\&=\Bigg(\left\lfloor \frac{305\times 1}{503}\right\rfloor+\left\lfloor \frac{305\times 502}{503}\right\rfloor \Bigg)+\cdots \Bigg(\left\lfloor \frac{305\times 251}{503}\right\rfloor+\left\lfloor\frac{305\times252}{503}\right\rfloor \Bigg) \\&=304 \times 251 \\&=76304 \end{align*}