smc

Let the first number have digits $\underline{a}$$\underline{b}$. Then, it equals $10a+b$, and the second number equals $10b+a$. Taking the difference of the squares of these two numbers and simplifying, we see that: $$\begin{array}{rl}(10a+b{)}^2-(10b+a{)}^2& =((10a+b)-(10b+a))((10a+b)+(10b+a))\\ & =(9a-9b)(11a+11b)\\ & =9\cdot 11(a-b)(a+b)\end{array}$$ Thus, the result in question is always divisible by $9$, $11$, $(a-b)$, and $(a+b)$. This fact eliminates all of our options except $\boxed{\text{the product of the digits}}$.