SUM TRICK

SUM TRICK, A MAGIC

Friends during a recent visit to a magic show, a person was asked to think of a



3-digit number, whose 100th digit was bigger than the rest numbers, for ex. 971 or



956 i.e. 9 > 5,6. And all the numbers should be distinct. Then the algorithm of the



magic went like this :

1- Take the number and from this number subtract the number obtained by



reversing the digits,

2- Now having this number add this to the number obtained by reversing its digits



(the new one, the above one in 2-part),

3- The number obtained will always be 1089!!! and i was shocked how could it be?



Funny isn't it but my question is that can anyone prove this with a mathematical



base. I tried it with a good success, its a challenge to all.

This example will clear any doubt in the algorithm:

Let a number be 871, the reverse number is 178. And now, 871-178=693.

Now reverse of 693 is 396. So 693+396=1089

It is believed that this trick was derived from a Russian Olympiad where students



were asked to prove this but only a handful of them could prove it.

Let the no. be ABC where A>B,C and all are distinct.Then,

ABC-CBA=(A-C-1)(9)(10+C-A) where the terms in brackets are the digits of the no.

Now,(A-C-1)(9)(10+C-A)+(10+C-A)(9)(A-C-1)=1089 since units digit is 9 and sum of digits in 10's place is 18,so,1 is added to 9,which is in 100's place.