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

PhDmessi wrote:Hi Mike,
what about following approach, if we presume your 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.

- My first attempt was to think, I need to find the largest number.
thus: 987
But then I realized, 987 - 789 = 198 (as you found out).

- Then I thought, ok if A=9, the C must be the smallest number, hence 1,so that we can subtract 1xx (three digit number) from the largest number. B does not matter, as the result is alwas the same.
So when I have:
981 - 189 = 792 I have the maximum possible number for ABC - CBA.

Correct? Or am I missing something here out?

Well, you are interpreting the question as finding the maximum difference, whereas I was looking at it as a counting problem --- what is the maximum number of possibilities for the difference? I have no doubt that 792 is the maximum difference, if that is what the question is asking.

You could also get that from the algebraic approach I used. Difference = (A - C)*99. Neither A nor C can be zero, so the maximum difference of two nonzero digits is 9 - 1 = 8, and 8*99 = 792.

The other possibilities are:

981 - 189 = 792
891 - 198 = 693
791 - 197 = 594
691 - 196 = 495
591 - 195 = 396
491 - 194 = 297
391 - 193 = 198
291 - 192 = 99

That's the whole field of possibilities, as an answer to the counting question.

Mike