I hope something good happens to you.
we denote that $\lfloor x \rfloor$ is floor function and $(a, b):=d$ is greatest common divisor between $a$, $b$.
we know that
$$\sum_{i=1}^{b-1} {\left\lfloor \frac{ai}{b}\right\rfloor}=\frac{(a-1)(b-1)}{2}+\frac{d-1}{2}$$
where positive integer $a$, $b$. But it depends on $d$.
Here is my question.
Is there any theorem calculating $\displaystyle\sum_{i=1}^{b-1} {\left\lfloor \frac{ai}{b}\right\rfloor}$ regardless of $d$?