# Why is $\left\lfloor\frac{\lfloor a\pi\rfloor}{a}\right\rfloor=3$ for $a>0$?

Why is this true? $$\left\lfloor\frac{\lfloor a\pi\rfloor}{a}\right\rfloor=3 \text{, for } a>0$$

I need this to solve the Ukraine Math Olymipiad 1999. "$$\lfloor\cdot\rfloor$$" indicates the floor function.

• Is $a$ supposed to be an integer? – Blue Jan 12 at 14:49

## 3 Answers

The equation is equivalent to

$$0\le\{\pi\}-\frac{\{a\pi\}}a<1.$$

The right inequality is always verified. The left one is certainly verified when

$$0\le\{\pi\}-\frac1a,$$ or $$a\ge\frac1{\{\pi\}},$$ which is a little more than $$7$$.

Assuming that $$a$$ is restricted to be a natural, it remains to try $$a=1,2,\cdots7$$. And as $$\{\pi\}<\dfrac17$$, all these values will work.

• @JohnDoe: ooops, yes of course. – Yves Daoust Jan 12 at 15:05
• Also, what do you mean when you say it suffices to try $a=\cdots$? Are you saying these are the only values of $a$ for which it works? – John Doe Jan 12 at 15:05
• @JohnDoe: no, these are the values for which the bounding argument doesn't work. – Yves Daoust Jan 12 at 15:06
• but $[6\times\pi]=[18.8...]=18$, $[18/6]==3$ does work. In fact it should always work for a natural number. I don't think those numbers are special – John Doe Jan 12 at 15:08
• @JohnDoe: I didn't say that these don't work, I said that they needed to be confirmed with another argument. – Yves Daoust Jan 12 at 15:09

It isn't true! a= 1/2 is a counter example. If a= 1/2 then $$a\pi$$ is 1.5707... and the floor or that is 1. Dividing that by 1/2 gives 2 which, of course, has floor 2, not 3.

• Perhaps the question intends that $a$ is an integer. – Blue Jan 12 at 14:47

Let $$a=b/\pi$$ for some $$b\in\Bbb R$$.

Then $$[a\pi]=[b]$$

$$[[b]/a]=\left[\pi\cdot\frac{[b]}{b}\right]$$ For this to equal 3, we need $$\frac{[b]}{b}\ge\frac3\pi\implies [b]\ge3b/\pi\approx 0.95 b$$

This fails for $$b\in\{0\}\cup\left(\frac{(n-1)\pi}{3},n\right)$$ for $$n\in\{1,2,\cdots,22\}$$, and thus for $$a\in\{0\}\cup\left(\frac{n-1}3,\frac n\pi\right)$$ For all other values of $$a$$, this holds.