A Divisibility Difference

Let the two digit number be $\overline{AB}=10A+B$ .Then the number forned by reversing its digits will be $\overline{BA}=10B+A$ .Their absolute difference would be: $\begin{aligned} |\overline{AB}-\overline{BA}|&=|(10A+B)-(10B+A)|\\ &=|9A-9B|=9|A-B|\end{aligned}$ Therefore the difference of a two digit number and the number formed by reversing its digits is always divisible by $\boxed{9}$

Side Note:

As it turns out,this is actually true for any number.That's because every number can be represented as $10A+B$ (Note that A may not be a single digit number this way,but it won't affect the result).The number formed by reversing its digits will be $10B+A$ .Their absolute difference is still $9|A-B|$ and therefore is still a multiple of 9.

I take the number 21 and reverse it as 12. Difference=9 So divisible by 9

$|\overline{ab}-\overline{ba}|=|10a+b-10b-a|=|9(a-b)|$ .Therefore it divisible by $\boxed{\large{9}}$