The floor is a function 2

Relevant wiki: Hermite's Identity

$S = \sum_{x = 2}^{2017} \sum_{y = 1}^{x - 1} \sum_{z = 0}^{y - 1} \left \lfloor \frac{2017y + xz}{xy} \right \rfloor$

$S = \sum_{x = 2}^{2017} \sum_{y = 1}^{x - 1} \sum_{z = 0}^{y - 1} \left \lfloor \frac{2017}{x} + \frac{z}{y} \right \rfloor$

$S = \sum_{x = 2}^{2017} \sum_{y = 1}^{x - 1} \left (\left \lfloor \frac{2017}{x} \right \rfloor + \left \lfloor \frac{2017}{x} + \frac{1}{y} \right \rfloor + \left \lfloor \frac{2017}{x} + \frac{2}{y} \right \rfloor + \ldots + \left \lfloor \frac{2017}{x} + \frac{y - 1}{y} \right \rfloor \right )$

Using Hermite's Identity ,

$\left \lfloor \frac{2017}{x} \right \rfloor + \left \lfloor \frac{2017}{x} + \frac{1}{y} \right \rfloor + \left \lfloor \frac{2017}{x} + \frac{2}{y} \right \rfloor + \ldots + \left \lfloor \frac{2017}{x} + \frac{y - 1}{y} \right \rfloor = \left \lfloor \frac{2017y}{x} \right \rfloor$

$S = \sum_{x = 2}^{2017} \sum_{y = 1}^{x - 1} \left \lfloor \frac{2017y}{x} \right \rfloor$

$S = \sum_{x = 2}^{2017} \left (\left \lfloor 2017 \times \frac{1}{x} \right \rfloor + \left \lfloor 2017 \times \frac{2}{x} \right \rfloor + \left \lfloor 2017 \times \frac{3}{x} \right \rfloor + \ldots + \left \lfloor 2017 \times \frac{x - 1}{x} \right \rfloor \right )$

Let $a$ be a positive integer less than $2017$ .

$\left \lfloor 2017 \times \frac{a}{x} \right \rfloor + \left \lfloor 2017 \times \frac{x - a}{x} \right \rfloor = 2017 \times \frac{a}{x} - \left \{ 2017 \times \frac{a}{x} \right \} + 2017 \times \frac{x - a}{x} - \left \{ 2017 \times \frac{x - a}{x} \right \} = 2017 - \left \{ 2017 \times \frac{a}{x} \right \} - \left \{ 2017 \times \frac{x - a}{x} \right \}$ .

Since the LHS is an integer, the RHS must also be an integer which forces the RHS to be equal to $2016$

There are $\left \lfloor \frac{x - 1}{2} \right \rfloor$ pairs that sums up to $2016$ and for even values of $x$ , there will be an extra $1008$ added to the sum.

$S = \sum_{x = 2}^{2017} \left ( 2016 \times \frac{x - 1}{2} \right )$

$S = 2016 \times \sum_{x = 1}^{2016} \frac{x}{2}$

$S = \frac{2016 \times 2016 \times 2017}{4}$

$\sqrt{S} = \frac{1008}{2} \times \sqrt{2017}$

$\left \lfloor \sqrt{S} \right \rfloor = \boxed{45270}$

Notation: $\lfloor \cdot \rfloor$ denotes the floor function .

and $\{ \cdot \}$ denotes the fractional part function .