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.
Why is $\left\lfloor\frac{\lfloor a\pi\rfloor}{a}\right\rfloor=3$ for $a>0$
ceiling-and-floor-functionspi
Best Answer
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.