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.

At  first I thought that looks too easy.  Then wait, "Multiple", "Factor"-- two different things.  Makes a big difference.  Still, I know it must be true, but to prove it, hmm.

Perhaps one approach may be to prove that there are a infinite numbers that do not contain the digits 0, 1, or 8.

Or could it be like "casting out nines".  I mean we know that every multiple will end in 5.  No I don't think that's it.

I suspect there are a couple of calc rules out there that will prove, but I don't know what they are.  Doubt I could ever get this one Federico, but I'll hang around to see your solution.  I just hope I can understand it. 

By the way, I like how you used decimal format in your definition.  I can easily imagine answers involving, well definitely not binary but maybe hex.

Edited on August 29, 2004, 2:51 pm

Posted by Bruce Brantley on 2004-08-29 14:37:38