CEMC Hypatia, 2015

The two-digit positive integers $AB$ and $BA$ equal $10A+B$ and $10B+A$, respectively.

Solving $AB-BA=72$, we get $(10A+B)-(10B+A)=72$ or $9A-9B=72$ and so $A-B=8$.

Since $A$ and $B$ are positive digits, then the only possibility for which $A-B=8$ occurs when $A=9$ and $B=1$. (Verify for yourself that this is indeed the only possibility.)

Therefore, the positive integer $AB$ is 91.
(We may check that $AB-BA=91-19=72$.)
The two-digit positive integers $MN$ and $NM$ equal $10M+N$ and $10N+M$, respectively.

Solving $MN-NM=80$, we get $(10M+N)-(10N+M)=80$ or $9M-9N=80$ and so $9(M-N)=80$.

Since $M$ and $N$ are positive digits, then $M-N$ is an integer and so $9(M-N)$ is a multiple of 9.

However, 80 is not a multiple of 9 and so $9(M-N)\neq80$.
Therefore, it is not possible that $MN-NM=80$.
The three-digit positive integers $PQR$ and $RQP$ equal $100P+10Q+R$ and $100R+10Q+P$, respectively.

Simplifying $PQR-RQP$, we get $(100P+10Q+R)-(100R+10Q+P)$ or $99P-99R$ or $99(P-R)$.

Since $P$ and $R$ are positive digits, the maximum possible value of $P-R$ is 8 (which occurs when $P$ is as large as possible and $R$ is as small as possible, or $P=9$ and $R=1$).

Since $\text{P gt;R}$, the minimum possible value of $P-R$ is 1 (which occurs when $P=9$ and $R=8$, for example).

That is, $1\leq P-R\leq 8$ and so there are exactly 8 possible integer values of $P-R$.

(Verify for yourself that there are values for $P$ and $R$ so that $P-R$ is equal to each of the integers from 1 to 8.)

Since $PQR-RQP=99(P-R)$ and there are exactly 8 possible values of $P-R$, then there are exactly 8 possible values of $PQR-RQP$.
(We note that the value of $PQR-RQP$ does not depend on the value of the digit $Q$.)