Will you help me in mathematics?

Question

I'm really bad at mathematics, and I really need help . . . so please help me. Well, first of all, I guess I could say I'm bad with algebra, or just variables. Like the following question.
- If the three digit numbers x and y are each composed of three distinct non-zero digits, what is the greatest possible value of x - y?

Please explain how to do the above problem, and the following one!
- In the addition problem below, if the letters A, B, and C represent different digits, what does each one equal?

AA + AB = CCC


Please help, and thanks ahead of time! (I really want to be as smart as all of you!)

Answer 1

You can write the numbers like this:

10A + A

10A + B

100C + 10C + C

since those letters represent the values of the ones, tens, and hundreds place. We know now A,B,and C are all digits (0 through 9).

We solve:

10A + A + 10A + B = 100C + 10C + C

simplify:

21A + B = 111C

Now, here's the trick. When you add two 2-digit numbers (here, AA and AB) you know they can't be more than 200. Therefore, C has to be 1 (the 100s place of CCC can only be 1).

Thus we know:

21A + B = 111.

Because of the 21 on A, we know that 21A contributes a ones place of A. That means the ones place of A+B is 1. So one of the following must be true:

A+B=1
A+B=11
A+B=21
A+B=31...etc

But since A and B are digits, they can't add up to more than 18. And clearly, A+B=1 means one of them is zero, and there's no way 21A+B = 111. So A+B=11.

That means though that now we have to resolve the tens place. A+B = 11 so we can simplify:

21A+B = 20A + (A+B) = 20A + 11 = 111
So 20A = 100

Thus:
A=5
B=6
C=1 (from before)

Check it

  55
+56
-----
111

It works. All set. I hope that made sense.

Answer 2

for the first problem
the greatest x is 987 and the smallest y is 123 so the greatest difference is 864
10A+A +10A+B=100C+10C+C
21A+B=111C
A at least must be 5 because for lower values 21A+B<100
with A=5
105+B=111C so C must be 1 and B =6
With A >5 there is no solution because B is not a digit