jmc

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 $\boxed{8}$