Math Magic demystified

Just returned home from the hospital. It has been a long stay(8 days), and its good to be home.

In the previous post, i showed a simple number trick. Let me now prove that it works for any three digit number.

Notice that all three digit numbers can be represented as 100a+10b+c where a,b and c are positive integers. This is simple stuff from primary school. So for eg, the number 684 would be 6(100)+8(10)+4. With that clarified, lets continue.

We start with the three digit number 100a+10b+c
Its reverse would be 100c+10b+a
their difference would be 100a+10b+c-100c-10b-a
which is equal to 100(a-c)+(c-a)=100(a-c)-(a-c)
=99(a-c)

since a-c can range only from 1 to 9 and is an integer, the difference of the two three digit numbers can only take 9 values . They are,

99 , 198 , 297 , 396 , 495 , 594 , 693 , 792 , 891

It is not difficult to see that the sum of the digits of all these numbers equal to 18, a pretty little coincidence, don't you think? But whats more beautiful is the technique used to prove the theorem. Originally, we had 990 numbers to test. But by using logic, we reduced that number to only 9 seperate cases. With todays technology, we could of course use a normal household computer to test all 990 numbers-it would probably finish the task before you can say "proven". However, there are certain theorems in mathematics that deal with an infinite number of numbers. And infinity is something not even the world's total processing power can attack. Besides, don't you think my short and simple proof is more elegant than the brute force of a computer?