Relevant wiki: Floor and Ceiling Functions - Problem Solving

Since 18 and 35 are coprime, then for integer $1\leq r \leq 34$ , the fractional part of $\dfrac{18r}{35}$ can take the values $\dfrac{1}{35}, \dfrac{2}{35} , \ldots, \dfrac{34}{35}$ .

By definition, $x - \{ x \} = \lfloor x \rfloor$ . We can rewrite the summation in question as

$\begin{aligned} \sum_{r=1}^{34} \left \lfloor \dfrac{18r}{35} \right \rfloor &=& \sum_{r=1}^{34} \left( \dfrac{18r}{35} - \left \{ \dfrac{18r}{35} \right \} \right) \\ &=& \sum_{r=1}^{34} \dfrac{18r}{35} - \sum_{r=1}^{34} \dfrac r{35} \\ &=& \sum_{r=1}^{34} \dfrac{17r}{35} = \dfrac{17}{35} \sum_{r=1}^{34} r \\ &=& \dfrac{17}{35} \cdot \dfrac{34(34+1)}2 = 17^2 = \boxed{289} . \\ \end{aligned}$

The second last step follows from the algebraic identity , $1 + 2 + 3 + \cdots + n = \dfrac {n(n+1)}2$ .

18r/35 = r/2 + r/70,
The contribution of r/70, towards sum of floor function would be zero for each of the number from 1 to 34 and, therefore, it can be ignored.
The contribution of r/2, towards sum of floor function would be reduced by 1/2, for each of the 17 odd numbers from 1 to 34.
Therefore, required sum;
= (1/2)(1+2+3+....+34) - 17/2,
= (1/2)(34×35/2) - 17/2 = (17/2)(35 -1),
= 17^2 = 289.

To get an intuition for the problem, try plugging in values and seeing what you get. When you plug in r = 1, you get 0. Then from there, r =2 and r = 3 both give you 1, r=4 and r=5 both give you 2, r=6 and r=7 both give you 3...etc.

Since this is a 33 term sequence between r=2 and r=34, the series can be thought of as

the sum from n=1 to n=16 of (2n) + 17 = 17*(16+1) = 17^2 = 289

include <iostream>

using namespace std;

int main() { int sum=0; for (int r=1;r<=34;r++) { sum+=(int)(18*r/35); } cout<<sum; }

My approach was as follows:

I noticed that each value of the fraction inside the floor function is slightly larger than $\frac{18r}{36}=\frac{r}{2}$ , which is much easier to sum:

$\displaystyle \sum_{r=1}^{34}\left \lfloor \dfrac{r}{2} \right \rfloor=0+2\times(1+2+\ldots +16)+17=17^2=\boxed{289}$ ,

which is the value asked for, since the accumulated excesses are insufficient to make any of the terms in the original sum jump to the next integer value.

We know that the first term in the sum drops out. We also know that every remaining term in the sum will appear twice with the exception of 17. Therefore, we multiply the sum of the natural numbers from 1 to 16 by 2 and add 17 for our final answer of 289.

I wrote this solution on a mobile device and didn't feel like using TeX.