Hide answer & extension
Call the digits of the starting number \(A\), \(B\) and \(C\).
If the number is a palindrome (if \(A=C\)) then the answer to the subtraction will be 0, so the answer is 0.
If the first and last digits differ one, we can consider the case \(A=C+1\). The same analysis will apply to \(C=A+1\).
$$100C+10B+A-(100A+10B+C)=99C-99A$$
$$=99A+99-99A$$
$$=99$$
Then, \(99+99=198\) so the answer is 198
If the first and last digits differ by at least two, we can consider the case \(A=C+n\), where \(2\leq n\leq 9\). The same analysis will apply to \(C=A+n\).
$$100C+10B+A-(100A+10B+C)=99C-99A$$
$$=99(A+n)-99A$$
$$=99n$$
This number will be 198, 297, 396, 495, 594, 693, 792, 891 or 990. In each of these numbers, the middle digit is 9 and the two other digits add up to 9, so they can be written \(100n+90+(9-n)\). If we add this to its reverse:
$$100n+90+(9-n)+100(9-n)+90+n$$
$$=100n+99-n+900-100n+90+n$$
$$=900+99+90+100n-100n+n-n$$
$$=1089$$
So the three possible answers are 0, 99 and 1089 which will occur when the first and list digits of the number differ by 0, 1 or more respectively.
Extension
Which answers are possible if the numbers are written in different bases?
Which answers are possible if four digit numbers are taken? Or five digit numbers?
