A Puzzle With 1089

Take a three digit number, with the first and third digits different.

           Example: 591

Reverse the digits.

           195

Subtract the smaller number from the larger one.

            591
          - 195
            396


Reverse the digits in this new number.

           693

Now add the two numbers.

             396
          + 693
           1089


Try it again with a different first number, say 589.

             985
           - 589
             396
          + 693
           1089


You will ALWAYS get 1089. Why??

Let the numbers be abc and cba, with abc > cba.
Notice that c will be LESS THAN a.
This means, when you subtract a from c, you will need to borrow:

                          b-1       10+c
              a            b            c
            - c            b            a
                                        10+c-a


Now you need to subtract b from (b-1), so you need to borrow from a:

               a-1        10+b-1         10+c
                a               b                c
            -   c               b                a    
             a-1-c     10+b-1-b     10+c-a   (*)


Now we simplify this 3-digit number. Notice that 10 + b - 1 - b = 9. So the three digits in (*) become:

           a-1-c     9     10+c-a

Now we add its reverse to it:

               a-1-c        9     10+c-a
           + 10+c-a      9      a-1-c   
                  9          18         9
              = 10          8          9
              = 1089


We have to carry the one from the second "digit", so we have

          1089

So the answer is always 1089!

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