The god of Number Theory

If you think about the floor function, repeated values start when

$\lfloor \frac{2018}{i} \rfloor = \lfloor \frac{2018}{i+1} \rfloor$

So, let's consider the bound

$1 \geq \frac{2018}{i} - \frac{2018}{i+1}$

This gives us an idea of where $\lfloor \frac{2018}{i} \rfloor$ might start having repeated values.

$(i+1)(i) - 2018 \geq 0$ , ( note: $i \in \mathbb{Z}$ )

$i \geq 45$

For $1 \leq i \leq 44$ , results are $44$ different discrete numbers, because consecutive results differ by more than 1.

For $45 \leq i \leq 2018$ , consecutive results will either differ by 1 or 0 (repeat), so unique results will decrease by 1 from $\lfloor \frac{2018}{45} \rfloor = 44$ to $\lfloor \frac{2018}{2018} \rfloor = 1$ . So, other $44$ discrete numbers are in the set.

$44+44 = 88$

Imagine the curve $y(x)=\frac{2018}{x}$ on real line. The derivative of the curve is $y'= -\frac{2018}{x^2}$ . We are gonna find the point, at which the derivative gets lower than one, in magnitude, by putting the derivative equal to minus one. That would be at $x=\sqrt{2018}\approx 44.9$ . When $1\leq i \leq 44$ , $\lfloor \frac{2018}{i}\rfloor$ takes $44$ different values, because the absolute value of the derivative is higher than $1$ and $\frac{2018}{i+1} < t-1$ , for $t \in \mathbb{R}$ , if $\frac{2018}{i} =t$ . For $45\leq i \leq 2018$ , the derivative would be less than one, in magnitude, therefore, all the values from $\lfloor \frac{2018}{45}\rfloor=44$ to $1$ , would be covered and there are $44$ of them. So, there would be a total of $44+44=88$ values that are produced.