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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.

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

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

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

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