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Not a Prime: Gee thanks.

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  • patience_and_fortitude
    Exactly what I said...within 5 min somebody would find one. Obviously somebody can write code really fast to check these kinds of weidisms... however, it does
    Message 1 of 10 , Oct 8, 2004
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      Exactly what I said...within 5 min somebody would find one.
      Obviously somebody can write code really fast to check these kinds of
      weidisms...

      however, it does hold true for 3 successive terms. but this is all
      too trivial to prove that any resulting number of a three term
      expansion is divisible by 3.

      188881888218883...
    • Jens Kruse Andersen
      ... Yes. The concatenation of 3n consecutive integers in ascending or descending order is divisible by 3. There are probably infinitely many primes in all
      Message 2 of 10 , Oct 8, 2004
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        patience_and_fortitude wrote:

        > however, it does hold true for 3 successive terms. but this is all
        > too trivial to prove that any resulting number of a three term
        > expansion is divisible by 3.

        Yes.
        The concatenation of 3n consecutive integers in ascending or descending order
        is divisible by 3.
        There are probably infinitely many primes in all other cases.

        See http://www.primepuzzles.net/puzzles/puzz_078.htm
        The hard cases are 22n terms, because they must cross a power of 10 to avoid
        the factor 11.
        The smallest number of terms without a known prime or prp is 44 descending
        terms.
        Can you succeed where I failed?
        My search was exhaustive to 22000 digits.
        Any prp will be unprovable today.

        --
        Jens Kruse Andersen
      • patience_and_fortitude
        As it turns out it s a little more general They don t even need to be consecutive pick a k and an n, and whala! it s divisible by 3 even if the digits of n
        Message 3 of 10 , Oct 8, 2004
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          As it turns out it's a little more general
          They don't even need to be consecutive
          pick a k and an n, and whala! it's divisible by 3 even if the digits
          of n increase mid number or if k>n...although proving this case might
          not be so trivial.

          (n)--(n+k)--(n+2k)

          k=10
          n=8
          81828

          k=2
          n=99
          99101103

          k=11
          n=9989
          99891000010011

          k=131
          n=2
          2133264

          k=12345
          n=123666
          123666136011148356

          (I'm fairly certain that's true)
          -Shawn

          --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
          <jens.k.a@g...> wrote:
          > patience_and_fortitude wrote:
          >
          > > however, it does hold true for 3 successive terms. but this is all
          > > too trivial to prove that any resulting number of a three term
          > > expansion is divisible by 3.
          >
          > Yes.
          > The concatenation of 3n consecutive integers in ascending or
          descending order
          > is divisible by 3.
          > There are probably infinitely many primes in all other cases.
          >
          > See http://www.primepuzzles.net/puzzles/puzz_078.htm
          > The hard cases are 22n terms, because they must cross a power of 10
          to avoid
          > the factor 11.
          > The smallest number of terms without a known prime or prp is 44
          descending
          > terms.
          > Can you succeed where I failed?
          > My search was exhaustive to 22000 digits.
          > Any prp will be unprovable today.
          >
          > --
          > Jens Kruse Andersen
        • patience_and_fortitude
          As it turns out it s a little more general They don t even need to be consecutive pick a k and an n, and whala! it s divisible by 3 even if the digits of n
          Message 4 of 10 , Oct 8, 2004
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            As it turns out it's a little more general
            They don't even need to be consecutive
            pick a k and an n, and whala! it's divisible by 3 even if the digits
            of n increase mid number or if k>n...although proving this case might
            not be so trivial.

            (n)--(n+k)--(n+2k)

            k=10
            n=8
            81828

            k=2
            n=99
            99101103

            k=11
            n=9989
            99891000010011

            k=131
            n=2
            2133264

            k=12345
            n=123666
            123666136011148356

            (I'm fairly certain that's true)
            -Shawn

            --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
            <jens.k.a@g...> wrote:
            > patience_and_fortitude wrote:
            >
            > > however, it does hold true for 3 successive terms. but this is all
            > > too trivial to prove that any resulting number of a three term
            > > expansion is divisible by 3.
            >
            > Yes.
            > The concatenation of 3n consecutive integers in ascending or
            descending order
            > is divisible by 3.
            > There are probably infinitely many primes in all other cases.
            >
            > See http://www.primepuzzles.net/puzzles/puzz_078.htm
            > The hard cases are 22n terms, because they must cross a power of 10
            to avoid
            > the factor 11.
            > The smallest number of terms without a known prime or prp is 44
            descending
            > terms.
            > Can you succeed where I failed?
            > My search was exhaustive to 22000 digits.
            > Any prp will be unprovable today.
            >
            > --
            > Jens Kruse Andersen
          • Jud McCranie
            ... You might be interested in Smarandache problems, a guy who posed a lot of problems similar to this one.
            Message 5 of 10 , Oct 8, 2004
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              At 09:42 PM 10/8/2004, patience_and_fortitude wrote:

              >As it turns out it's a little more general
              >They don't even need to be consecutive
              >pick a k and an n, and whala! it's divisible by 3 even if the digits
              >of n increase mid number or if k>n...although proving this case might
              >not be so trivial.
              >
              >(n)--(n+k)--(n+2k)

              You might be interested in Smarandache problems, a guy who posed a lot of
              problems similar to this one.
            • mikeoakes2@aol.com
              In a message dated 09/10/2004 02:44:15 GMT Daylight Time, shawns_email_address@yahoo.com writes: As it turns out it s a little more general They don t even
              Message 6 of 10 , Oct 9, 2004
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                In a message dated 09/10/2004 02:44:15 GMT Daylight Time,
                shawns_email_address@... writes:



                As it turns out it's a little more general
                They don't even need to be consecutive
                pick a k and an n, and whala! it's divisible by 3 even if the digits
                of n increase mid number or if k>n...although proving this case might
                not be so trivial.

                (n)--(n+k)--(n+2k)

                k=10
                n=8
                81828

                [snip]



                Have you not heard of the elementary-school result that if the sum of the
                digits of a number is divisible by 3 then the number itself is divisible by 3?
                (Ditto 9.)

                n + (n+k) + (n+2k) = 3*(n+k).

                whala!

                -Mike Oakes


                [Non-text portions of this message have been removed]
              • Jud McCranie
                ... I thought he was stringing the decimal digits of the terms together. What happens when n+2k has more digits than n?
                Message 7 of 10 , Oct 9, 2004
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                  At 05:12 AM 10/9/2004, mikeoakes2@... 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?
                • mikeoakes2@aol.com
                  In a message dated 09/10/2004 16:18:39 GMT Daylight Time, j.mccranie@adelphia.net writes: ... I thought he was stringing the decimal digits of the terms
                  Message 8 of 10 , Oct 9, 2004
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                    In a message dated 09/10/2004 16:18:39 GMT Daylight Time,
                    j.mccranie@... writes:

                    At 05:12 AM 10/9/2004, mikeoakes2@... 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]
                  • patience_and_fortitude
                    Thanks David. This is where it was ultimately going I m sure... -Shawn ... together. What ... 3 after
                    Message 9 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]
                    • patience_and_fortitude
                      Thanks David. This is where it was ultimately going I m sure... -Shawn ... together. What ... 3 after
                      Message 10 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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