Thanks David.

This is where it was ultimately going I'm sure...

-Shawn

--- 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 primenumbers@yahoogroups.com, 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

together. What

> 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

3" after

> my equation.

>

> -Mike Oakes

>

>

> [Non-text portions of this message have been removed]