Floor, ceil problem

Question

Edit-For what values of $k$, -$2$+ $\left\lceil\frac{(6k+1)\pi }{6}\right\rceil=\left\lfloor\frac{(6k-1) \pi}{6}\right\rfloor$ where $k$ is positive integer greater than equal to one?

Answer 1

The claim is wrong. When $k=39$ we have $$\left\lceil{(6k-1)\pi\over6}\right\rceil=\lceil 121.9985\rceil=122,\qquad \left\lfloor{(6k+1)\pi\over6}\right\rfloor=\lfloor 123.0457\rfloor=123\ .$$

Answer 2

You should convince yourself that you are looking for precisely the $k$ s.t. $\left(\frac{(6k-1)\pi}{6}, \frac{(6k+1)\pi}{6}\right) \cap \mathbb{Z} \neq \emptyset$. Then this happens iff $((6k-1) \pi, (6k+1)\pi) \cap 6\mathbb{Z} \neq \emptyset$.