As part of a problem I'm solving, I stumbled across the following system of integer inequalities, for which I cannot seem to find a mathematical method to solve it. The system is as follows: $$\left\{\begin{matrix} B\geq 1\\ B\geq -\frac{5}{4}k\\ B\leq -\frac{4}{3}k \end{matrix}\right. \quad\text{or put differently: } \left\{\begin{matrix} B\geq 1\\ k\geq -\frac{4}{5}B\\ k\leq -\frac{3}{4}B \end{matrix}\right.$$ We need to find $B\in\mathbb{Z}$ and $k$ can be any integer. E.g. if $B=1$ the system is not valid since we cannot find a $k$ that satisfies all equations. It is not hard to see that $B \geq 12$ is a solution for this problem: $12$ is the first positive integer for which we can find a valid $k$ (in this case $-9$). Any integer $B$ greater than $20$ is also a solution since then the difference between $-\frac{3}{4}B$ and $-\frac{4}{5}B$ is greater than $1$, so we can always find at least one integer $k$ that satisfies the equations.
EDIT: I did not explain where this problem comes from, but $B=4,5,8,9,10$ are also solutions. Only for $B\geq 12$ are all consecutive positive integers solutions. That is actually what I'm interested in.
Showing that $B\geq 20$ are consecutive solutions is easy (as explained above), but how can you show that $12\leq B < 20$ are also consecutive solutions? This is, of course, a toy problem with rather simple fractions, so it would be nice if someone could come up with a rigorous plan to tackle this in case the system gets more complicated. It reminds me somewhat of integer programming... Thanks!