# A theorem about the Floor function

Prove that the amount of integers between $1,\ldots,n$ which are divisible by $1\le k\le n$ is $\left\lfloor\dfrac{n}{k}\right\rfloor$ ?

• I think "divided" should be "divisible" – steven gregory Sep 6 '17 at 15:25

## 2 Answers

There are certainly the $\left\lfloor \frac nk\right\rfloor$ integers $k,2 k,\cdots, \left\lfloor \frac nk\right\rfloor k$.

On the other hand, if $ku\le n$, then $u\le\frac nk$ and, since $u$ is an integer, by definition $u\le\left\lfloor\frac nk\right\rfloor$. So, there are at most $\left\lfloor \frac nk\right\rfloor$ possible candidates for $u$.

• Sorry, can you start over again, and rigorously? – יהודה שמחה ולדמן Sep 6 '17 at 15:33
• I am not doing that. – user228113 Sep 6 '17 at 15:38
• Did I say something wrong? – יהודה שמחה ולדמן Sep 6 '17 at 15:46
• No, you did not. You asked me to re-do from scratch something that I had already done. And I shan't do it. – user228113 Sep 6 '17 at 15:49

The numbers from $1,2,\ldots,n$ which are divisible by $k$ are $k,2k,\ldots$. If the last term of this list be $mk$ then $mk\le n<(m+1)k$ i.e. $m$ is a integer such that $m\le \dfrac{n}{k} <m+1\Rightarrow m=\left\lfloor\dfrac{n}{k}\right\rfloor$. So there are $\left\lfloor\dfrac{n}{k}\right\rfloor$ numbers in the list.

• Thank you so much! I like your way. However, a few minutes before I saw this comment, it finally hit me from the definition itself $\left\lfloor\dfrac{n}{k}\right\rfloor\le\dfrac{n}{k}$ . – יהודה שמחה ולדמן Sep 6 '17 at 17:00