Prove that for any positive integer n, there exists at least one multiple of 5^n that doesn't have any perfect cube digits (0, 1, or 8) in its decimal representation.

This can be solved in a similar fashion as "Spirit of '76 with no products". As an algorithm:

Set K=5^N


Set L=K


Set I=1


while I isn't greater than N


   if the I-th digit from the right of K is 0, 1, 8


      add L to K


   add 1 to I


   multiply L by 10


the answer is the N rightmost digits of K

Posted by e.g. on 2004-08-30 08:35:32