Half of my 2-digit number N is also a 2-digit number,
one of its digit being the sum of N's digit, the other their difference.
All 4 digits are different.

Find my number.

By the way, I notice that the algebraic approach demonstrates that there is only one solution.  But that approach never took advantage of all 4 digits being different.  Therefore, the fact that all 4 digits are different is not needed to solve the problem.

Posted by Steve Herman on 2018-10-25 09:45:15