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More Generally: n*10^a + (n+k)*10^b + (n+2*k)*10^c = 0 mod 3

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  • patience_and_fortitude
    Thanks David. This is where it was ultimately going I m sure... -Shawn ... together. What ... 3 after
    Message 1 of 10 , Oct 9, 2004
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      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]
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