well...its simple as given in the article.. note that the numbers are all decimal numbers (not binary..in case misled by just 0 and 1 in the number) therefore if you have have k copies of 001 and k%3 == 0, then the sum of the digits is divisible by 3 hence the number is divisible by 3 - The Standard Test(The only digit is 1 and it occurs k times).
For the numbers such that k%3 <> 0, observe that the number mod k 9s has the value of k1s. Since k9s is also divisible by k1s the original number k*001* should also be divisible by k1s. This property holds good for all k%3<>0. --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
