Number of Multiples less than Given Number

Theorem
Let $m, n \in \N_{\ge 1}$.

The number of multiples of $m$ not greater than $n$ is given by:


 * $q = \left \lfloor {\dfrac n m}\right \rfloor$

where $\left \lfloor {\cdot} \right \rfloor$ is the floor function

Proof
By the Division Theorem:


 * $(1): \quad n = qm + r$

where $0 \le r < q$.

As $r < q$, it follows that the greatest multiple of $m$ up to $n$ is $q m$.

So all the multiples of $m$ up to $n$ are:
 * $m, 2m, 3m, \ldots, qm$

Dividing both sides of $(1)$ by $q$:


 * $(2): \quad \dfrac n m = q + \dfrac r m$

Taking the floor of $(2)$:


 * $\left \lfloor {\dfrac n m} \right \rfloor = \left \lfloor{q + \dfrac r m}\right \rfloor$

But as $0 \le \dfrac r m < 1$:
 * $\left \lfloor{q + \dfrac r m}\right \rfloor = q$

Recall all the multiples of $m$ up to $n$ are $m, 2m, 3m, \ldots, qm$.

It follows that the number of multiples of $m$ up to $n$ is:
 * $q = \left \lfloor {\dfrac n m} \right \rfloor$

Also see

 * De Polignac's Formula