Hi Cyril, The fact that there exists a period for your procedure (whatever the number of digits might be) is rather easy. Indeed, say you start with a n-digit number than after just a single step you get another n-digit number and so on. As there are only 10^n n-digit numbers you always obtain a period of at most 10^n numbers. You can get a smaller upper bound for the period because you always re-order the digits in a decending order, thus the number of ordered n-digit numbers is an upper bound for the period. Benny -------------------------------------------------------------------- CYRIL WROTE: >I''m searching a prove of a little problem. >You take a four digit number where not all digits are equal such 5957 and >reorder the digits >such that the biggest digit is at the first place, the second at the second >place etc. >Then subtract the smallest possible reoredering from the the other number >and restart the process. >As result you will get 6174. > >Example: 5957 > >9755-5579=4176; 7641-1467=6174; 7641-1467=6174 > >If you take a five digit number, the result will be a period of 74943, >62964, 71973, 83952. >With six digits it will be a period too. >Have anyone got a prove tor that? ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
