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Re: Multiplying Primes and Numbers in General

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  • djbroadhurst
    ... Not at my high school, where c = 6*a*b + a*b1 + b*a1 David
    Message 1 of 7 , Jun 5, 2013
      --- In primenumbers@yahoogroups.com,
      "mikeoakes2" <mikeoakes2@...> wrote:

      > (6*a+a1)*(6*b+b1) = 6*c+c1,
      > where c = 6*a*b+a1+b1,
      > and c1 = a1*b1

      Not at my high school, where
      c = 6*a*b + a*b1 + b*a1

      David
    • Kermit Rose
      Yes. I understand your concept. I have worked with those equations also. (6 a -1)(6b - 1) = (36 a b - 6 a - 6 b + 1) = 6 ( 6 ab - a - b) + 1 (6 a - 1) (6 b +
      Message 2 of 7 , Jun 19, 2013
        Yes.

        I understand your concept.

        I have worked with those equations also.

        (6 a -1)(6b - 1) = (36 a b - 6 a - 6 b + 1) = 6 ( 6 ab - a - b) + 1

        (6 a - 1) (6 b + 1) = 36 a b + 6 a - 6 b - 1 = 6 ( 6 a b + a - b) - 1

        (6 a + 1) (6 b - 1) = 36 a b - 6 a + 6 b - 1 = 6 ( 6 a b - a + b) - 1

        (6 a + 1) (6 b + 1) = 36 a b + 6 a + 6 b + 1 = 6 ( 6 a b + a + b) + 1


        Kermit
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