# four digit numbers

palomine
Posers and Puzzles 13 Dec '03 21:20
1. 13 Dec '03 21:201 edit
take any four digit number where all the digits are not the same. rearrange them to make the largest # possible. Rearrange them again to make the smallest # possible. Subtract the smaller # from the larger #. Take the result and do it all over again. repeat this process until you hit 6714, you should get stuck there. if you ever get a result less than four digits put a zero in front of it. (999 would be 0999).

Ex. 2372

7322 - 2237 = 5085

8500 - 5058 = 3442

4432 - 2344 = 2088

8802 - 2088 = 6714

Why is this?
2. TheMaster37
Kupikupopo!
14 Dec '03 10:58
Originally posted by palomine
take any four digit number where all the digits are not the same. rearrange them to make the largest # possible. Rearrange them again to make the smallest # possible. Subtract the smaller # from the larger #. Take the result and do it all over again. repeat this process until you hit 6714, you should get stuck there. if you ever get a result less than fo ...[text shortened]... - 2237 = 5085

8500 - 5058 = 3442

4432 - 2344 = 2088

8802 - 2088 = 6714

Why is this?
A guess i haven't proved yet; because of these calculations you cycle through all numbers there are. As soon as you have the 1467 you end up in a cycle;

7641-1467=6174
3. Fiathahel
Artist in Drawing
15 Dec '03 15:24
I don't why it is true, but it is easy to prove that all numbers you get are divisible by 9. This is cause the sum of the digits modulo 9 equals the number itslef modulo 9. And the first are the same for both numbers so the difference always equas 0 modulo 9.
Perhaps this might help a bit.

Steffin