If $pq$ is a two-digit number, then $pq-qp$ will be completely divisible by:

A two-digit number $pq$ can be mathematically represented in expanded place value form as:
$$pq=10p+q$$
Similarly, reversing its digits gives $qp$:
$$qp=10q+p$$
Let's find the difference:
$$pq-qp=(10p+q)-(10q+p)$$
$$pq-qp=9p-9q=9(p-q)$$
Since the difference is a multiple of $9$, it will always be completely divisible by $9$.