Sum of $34$ Terms Involving Floor Function

Question

Find the value of $$\left\lfloor \frac{18}{35}\right\rfloor +\left\lfloor \frac{18\cdot 2}{35}\right\rfloor +\left\lfloor \frac{18\cdot 3}{35}\right\rfloor +\cdots \cdots +\left\lfloor \frac{18\cdot 34}{35}\right\rfloor.$$

I tried the following:

Let $$S=\left\lfloor \frac{18}{35}\right\rfloor +\left\lfloor \frac{18\cdot 2}{35}\right\rfloor +\left\lfloor \frac{18\cdot 3}{35}\right\rfloor +\cdots \cdots +\left\lfloor \frac{18\cdot 34}{35}\right\rfloor.$$

Then write in the reverse order $$S=\left\lfloor \frac{18\cdot 34}{35}\right\rfloor +\left\lfloor \frac{18\cdot 33}{35}\right\rfloor +\left\lfloor \frac{18\cdot 32}{35}\right\rfloor +\cdots \cdots +\left\lfloor \frac{18}{35}\right\rfloor.$$

Now adding these two equation. I am not getting what to do; help me please.

Answer 1

Hint: Use $$\left\lfloor \frac{18 \times 2n}{35}\right\rfloor = \left\lfloor \frac{(1+35)\times n}{35}\right\rfloor=\left\lfloor n + \frac{n}{35}\right\rfloor=n$$ for $0 \le n<35$.

Also, $$\left\lfloor \frac{18 \times (2n+1)}{35}\right\rfloor = \left\lfloor \frac{(1+35)\times n +18}{35}\right\rfloor=\left\lfloor n + \frac{n+18}{35}\right\rfloor=n$$ for $0 \le n < 17$.