Re: [math-fun] n and n^3 have all digits odd

On Thursday 19 April 2012 03:57:09 Victor Miller wrote:

I'm doing some calculations pertaining to this. For all positive integers k define the set

S_k := { 1 <= n < 10^k | OddDigits(n) & OddDigits((n^3) mod 10^k) }

It looks like (log # S_k)/k --> 1 (where log is the natural log). Notice that 10^k-1 is always in S_k.

What happens in other bases? (6 and 14 seem like the closest counterparts of 10.) -- g