Hi, there. I'll give my 2 cents on this.
First of all, a very GMAT like question is the question:
A, B and C are 3 different positive integers. ABC is a three digit integer. What is the maximum number of possibilities for distinct values of ABC?
That's a relatively straightforward question involving the Fundamental Principle of Counting.
A can't be zero, so 9 possibilities for A. After one digit is chosen for A, 9 possibilities left for B. After one digit is chosen for A, and another for B, 8 possibilities left for C.
total # of possibilities are = 9*9*8 = 648
That's the number of three-digit numbers with three distinct digits.
Now, from this, which is very GMAT like, to your first question:
A, B and C are 3 different positive integers. ABC is a three digit integer. What is the maximum possible number for ABC - CBA.
I notice that if we find one difference, then increase all the digits by the same amount, then we get the same difference. For example
321-123 = 198
432-234 = 198
543-345 = 198
654-456 = 198
765-567 = 198
876-678 = 198
987-789 = 198
So, in this instance, there are 7 pairs that have the same difference. In another example:
381-183 = 198
492-294 = 198
Strikingly, the same difference as the first list.
Playing around with my calculator, I am finding that the differences can only be the positive or negative of one from the following list: {99, 198, 297, 396, 495, 594. 693, 792}. These are all multiples of 99.
Let N1 = ABC, and N2 = CBA, in "digits form". Then
N1 = A*100 + B*10 + C
N2 = C*100 + B*10 + A
N1 - N2 = (A - C)*100 + (C - A) = (A - C)*100 - (A - C) = (A - C)*99
Therefore, the difference always has to be a multiple of 99. All very interesting, but I'd say considerably harder than anything the GMAT would ask.
Here's a somewhat more GMAT-like question on digit combinations:
https://gmat.magoosh.com/questions/849
When you submit your answer to that question, it will be followed by a video explanation.
Let me know if anyone reading this has any questions about what I've said.
Mike
