Quantity of Positive Integers Divisible by Particular Integer

Theorem
Let $d$ be a positive integer.

Let $x \ge 1$ be a real number.

Then:


 * $\ds \sum_{n \le x, \, d \divides n} 1 = \floor {\frac x d}$

That is:


 * there are $\floor {\dfrac x d}$ natural numbers less than or equal to $x$ that are divisible by $d$.

Proof
Consider the sum:


 * $\ds \sum_{n \le x, \, d \divides n} 1$

Note that a natural number $n \le x$ is divisible by $d$ :


 * there exists a natural number $k$ such that $n = d k$.

So we are counting the natural numbers $k$ such that $d k \le x$.

That is, the natural numbers $k$ such that:


 * $k \le \dfrac x d$

So: