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

Question

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.

Answer 1

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 $$\sqrt{200\pi + \pi^2} - \pi<a<\sqrt{200\pi}$$

or approximately $$22.1<a<25.1$$

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.

Answer 2

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 $a=25$