How to solve this for every integer n, ⌊⌊n/2⌋/3⌋=⌊n/6⌋?

Question

The hint that is given for this question is Start by dividing the proof into two cases: n is even and n is odd. In case n is odd, use the quotientremainder theorem with divisor equal to 6 to divide into three cases: n 5 6k11, n 5 6k13, and n 5 6k15 for some integer k. You will need to consider a total of four cases.

Answer 1

Here is one of the four cases (specifically, one of the three cases where $n$ is odd):

Suppose $n = 6k + 5$, where $k$ is an integer. Then the LHS evaluates to \begin{align*} \left\lfloor \frac{ \lfloor n/2 \rfloor }{3} \right\rfloor &= \left\lfloor \frac{ \lfloor (6k+5)/2 \rfloor }{3} \right\rfloor = \left\lfloor \frac{ \lfloor 3k + 5/2 \rfloor }{3} \right\rfloor \\ &= \left\lfloor \frac{ \lfloor (3k + 2) + 1/2 \rfloor }{3} \right\rfloor \left\lfloor \frac{ 3k + 2 }{3} \right\rfloor = \left\lfloor k + \frac{2}{3} \right\rfloor = k \end{align*} while the RHS is $$\left\lfloor \frac{n}{6} \right\rfloor = \left\lfloor \frac{6k+5}{6} \right\rfloor = \left\lfloor k + \frac{5}{6} \right\rfloor = k.$$ So both sides are equal. Now it just remains to check this for all other cases, which are fairly similar.