More Generally: n*10^a + (n+k)*10^b + (n+2*k)*10^c = 0 mod 3
- Thanks David.
This is where it was ultimately going I'm sure...
--- David Broadhurst <D.Broadhurst@...> wrote:
> More generally
> n*10^a + (n+k)*10^b + (n+2*k)*10^c = 0 mod 3
> for all integers n,k,a,b,c,
> since 10 = 1 mod 3
--- In firstname.lastname@example.org, mikeoakes2@a... wrote:
> In a message dated 09/10/2004 16:18:39 GMT Daylight Time,
> j.mccranie@a... writes:
> At 05:12 AM 10/9/2004, mikeoakes2@a... wrote:
> >n + (n+k) + (n+2k) = 3*(n+k).
> I thought he was stringing the decimal digits of the terms
> happens when n+2k has more digits than n?
> It doesn't matter!
> xyz00000000000 = xyz mod 3.
> There was an implied (and for clarity I should have written) a "mod
> my equation.
> -Mike Oakes
> [Non-text portions of this message have been removed]