# How to solve function with floor: $a\left(\big \lfloor\frac {a} {2\pi}\big \rfloor + 1\right) = 100$

I've never worked with equations that contains floor. I wonder how would you solve $$a\left(\big \lfloor\frac {a} {2\pi}\big \rfloor + 1\right) = 100$$

I looked it up in the internet but I don't get it. So what I found was that, lets say $$\lfloor\frac {a} {2\pi}\rfloor$$ is equal to $$n$$ and $$n \le \frac {a} {2\pi} < n+1$$ and also $$a(n+1) = 100$$ but I can't continue from here. is there some way to calculate floor for general $$a$$ because I'm looking for value of $$a$$

$$a$$ is an integer

I got that I can try different variations like (100, 1), (50, 2), (25, 4) and (20, 5) and only 25 works but because I haven't solved an equation with floor function I want to know the techniques to solve it more mathematically and how it's done in general.

• Is $a$ an integer? A real? It makes big difference
– Zima
Commented Jun 14, 2023 at 10:31
• a is an integer Commented Jun 14, 2023 at 10:32
• Then I suggest you edit the post
– Zima
Commented Jun 14, 2023 at 10:33
• Then just try out all $a\mid100$. There's not that many. Commented Jun 14, 2023 at 10:33
• Note: your inequalities are not correct. If $n=\big \lfloor \frac a{2\pi} \big \rfloor$ then $n≤\frac a{2\pi}<n+1$.
– lulu
Commented Jun 14, 2023 at 10:36

We have $$\frac{a}{2\pi} < \left\lfloor \frac{a}{2\pi}\right\rfloor+1<\frac{a}{2\pi}+1$$

If $$a>0$$, then multiplying everything by $$a$$ gives $$\frac{a^2}{2\pi} < 100 < \frac{a^2}{2\pi}+a$$

The two inequalities can be solved to give $$\sqrt{200\pi + \pi^2} - \pi

or approximately $$22.1

We can either check the three integers in this range or just note the only factor of $$100$$ here is $$a=25$$.

If $$a<0$$, the inequalities become $$\frac{a^2}{2\pi} > 100 > \frac{a^2}{2\pi}+a$$

and in a similar way to the above, we can check that there are no negative solutions.

• ok that's interesting thanks a lot I got it now Commented Jun 14, 2023 at 11:00
• Great. I think it was just a different way of setting out the inequalities, but glad it helped. Commented Jun 14, 2023 at 11:05

One possibility, though very similar to what you proposed but presented differently, is to write $$a=2n\pi+\theta$$ with $$\theta\in[0,2\pi)$$

The equation becomes $$\ a(n+1)=100\iff(n+\frac{\theta}{2\pi})(n+1)=\frac{100}{2\pi}\$$ and we can bound: $$n(n+1)\le\frac{100}{2\pi}<(n+1)^2$$

whose only solution is $$n=3$$

$$\frac{100}{2\pi}\approx 15.91$$ gives interval $$[12,16]$$ for $$n=3$$, and negatives don't work, as it's either $$[9,12]$$ or $$[16,20]$$

Finally substituting $$n$$ gives $$4a=100$$ or $$a=25$$