How to show that $\left\lfloor{N/3}\right\rfloor < \left\lfloor{\left({N + 3}\right)/p}\right\rfloor$ has no solution for $p \ge 5$ where $N \ge p$

Question

I can see this numerically however I am having difficulty proving this by number theory methods. Both $N$ and $p$ are positive integers.

Answer 1

If $\lfloor \frac{N}{3} \rfloor < \lfloor \frac{N+3}{p} \rfloor $, then $\frac{N}{3} < \frac{N+3}{p} $ so that $pN < 3(N+3)$ or $N(p-3) < 9$.

But $N \ge p \ge 5$ so that $N(p-3) \ge 5\cdot 2 = 10 $, which contradicts this.

This was surprisingly easy.