$$\rm Method$$ $$1$$:  Let the hundreds, tens, and units digits of the original three-digit number be $$a$$, $$b$$, and $$c$$, respectively. We are given that $$a=c+2$$. The original three-digit number is equal to $$100a+10b+c=100(c+2)+10b+c=101c+10b+200$$. The hundreds, tens, and units digits of the reversed three-digit number are $$c$$, $$b$$, and $$a$$, respectively. This number is equal to $$100c+10b+a=100c+10b+(c+2)=101c+10b+2$$. Subtracting this expression from the expression for the original number, we get $$(101c+10b+200)-(101c+10b+2)=198$$ . Thus, the units digit in the final result is $$8$$.  $$\rm Method$$ $$2$$:  The result must hold for any three-digit number with its hundreds digit being $$2$$ more than the units digit. $$301$$ is such a number. Evaluating, we get $$301-103=198$$. Thus, the units digit in the final result is $$8$$.  ($$2010$$ AMC $$8$$ Problem, Question \#$$22$$) 
The hundreds digit of a three-digit number is $$2$$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result? 
Options:
(1) $$0$$ 
(2) $$2$$ 
(3) $$4$$ 
(4) $$6$$ 
(5) $$8$$ 