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.

Ok,  I see how you can remove any cube digit at the Mth position by adding n*10^m.  Because n*10^m is a multiple of 5 and the original was a multiple of 5 their sum must also be a multiple of 5.  And if you do this for every cube digit you eventually will have a multiple without any of your cube digits.  This seems to be proof enough to me. 

You lost me at X.10^n+Y.  I would like to think that because I'm just getting to Derivative Composition or the Chain Rule explains why I don't get it, but your proof appears to be pure algebra.  I don't know if it would be worth your time to expand on this a little.  I don't want to impose, but this problem really captured my interest.   

Posted by Bruce Brantley on 2004-10-16 10:11:34