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Re: [PrimeNumbers] Re: Multiplying Primes and Numbers in General

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  • G Fischer
    Hit send again and realized I misspoke below for the 6N+1 multiplied by 6N-1 numbers... c = 6ab + a - b This formula will show differently since a and b can be
    Message 1 of 7 , Jun 4 10:49 PM
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      Hit send again and realized I misspoke below for the 6N+1 multiplied by 6N-1
      numbers...
      c = 6ab + a - b
      This formula will show differently since a and b can be different... when
      the numbers are interchanged.

      As an example: 59*67 = 3953
      The formula for c in this case is: c = 6ab + a - b
      a = 10
      b = 11
      c = 659
      c = 6*10*11 - 10 + 11
      c = 660 + 1
      c = 661

      and flopped:
      a = 11, b = 10, c = 659
      c = 6*10*11 - 11 + 10
      c = 660 - 1
      c = 659

      Perhaps a rule for whichever number is 6N-1 is added and 6N+1 is subtracted?

      I don't know...it's getting complicated and interesting...as all prime
      numbers are.

      Thanks again all.

      Cheers,
      Geoff

      -----Original Message-----
      From: G Fischer
      Sent: Tuesday, June 04, 2013 10:33 PM
      To: primenumbers@yahoogroups.com ; mikeoakes2
      Subject: Re: [PrimeNumbers] Re: Multiplying Primes and Numbers in General

      Mike,

      Yep...Should have double checked and should have provided an example. I've
      probably now gone too far below...but hey, I was having fun going through
      the examples...

      We'll take your original 5*11=55
      5 is expressed as 6a-1 or 5 = 6(1) - 1 ... so a = 1
      11 is expressed as 6b-1 or 11 = 6(2) - 1 ... so b = 2
      55 is expressed as 6c + 1 (where the error was...when two 6N-1 numbers are
      multiplied, they always result in a 6N+1) or in this case 55 = 6(9)+1 ... c
      = 9
      a = 1
      b = 2
      c = 9

      Correct formula for two 6N-1 numbers being multiplied is:
      (6a-1)(6b-1) = 6c+1
      Then in this case:
      c = 6ab - b - a
      c = 6*1*2 - 2 - 1
      c = 12 - 3
      c = 9

      Regarding your comment: "Anyway, a formula of that kind can't be right, as
      interchanging a and b gives a different answer"
      Trying out 11*5 = 55
      a = 2
      b = 1
      c = 9

      c = 6ab - b - a
      c = 6*2*1 - 1 - 2
      c = 12 - 3
      c = 9

      Seems to work with either order...

      Take 11*17=187
      a = 2
      b = 3
      c = 31

      c = 6ab - b - a
      c = 6*2*3 - 3 - 2
      c = 36 - 5
      c = 31

      31*6 + 1 = 186 + 1 = 187

      Regarding your comment: "but multiplication is commutative"
      I'm going to have to look that up. I know it sounds basic...but that's
      where I'm at.

      If I wanted to do it with two 6N+1 numbers being multiplied, here is an
      example:
      7*13=91
      (6a+1)(6b+1)=(6c+1) -- since when 6N+1 numbers are multiplied together they
      also always equal a 6N+1
      c = 6ab + a + b
      a = 1, b = 2, c = 15
      c = 6*1*2 + 1 + 2
      c = 12 + 3
      c = 15

      Since both numbers are multiplied by 6 and then both added or subtracted, it
      doesn't matter which number is first when it's multiplied.

      And finally, when a 6N+1 number is multiplied by a 6N-1 it always results in
      a 6N-1 ...
      As an example: 59*67 = 3953
      The formula for c in this case is: c = 6ab + a - b
      a = 10
      b = 11
      c = 659
      c = 6*10*11 - 10 + 11
      c = 660 - 1
      c = 659

      Then 5*7 = 35
      a = 1
      b = 1
      c = 6
      c = 6*1*1 - 1 + 1
      c = 6 - 0
      c = 6

      Since a and b are multiplied by 6 and both are added and subtracted it
      doesn't matter which goes first.

      Regarding your final comment:
      Another general point: I should have thought that whether or not the numbers
      are /prime/ must be irrelevant in equations of the kind you are considering,
      no?
      Try some non-prime numbers.

      -- You are correct in that this should apply to all numbers, not just
      primes. But I haven't gotten that far.

      The bottom line to all the examples is that I'm wondering where to go to
      find out where this type of equation was discovered, by whom...applications
      used...etc.
      If given a multiplied number, can these equations be used to determine what
      numbers were used to make it? ... probably not, but could be fun.

      It's something cool that I've found on my own and I'm interested in what
      others have done with it.

      Sorry about the bad math...should have done better.

      Thanks for your help.

      Cheers,
      Geoff

      -----Original Message-----
      From: mikeoakes2
      Sent: Tuesday, June 04, 2013 12:04 AM
      To: primenumbers@yahoogroups.com
      Subject: [PrimeNumbers] Re: Multiplying Primes and Numbers in General



      --- In primenumbers@yahoogroups.com, "golfbum71" <golfbum71@...> wrote:
      >
      > I'm trying to find more resources to help verify something I've noticed.
      > You guys are all math geniuses (okay most of you are...but I won't say
      > who...) and I'm hoping someone can point me in the right direction.
      >
      > When multiplying two prime numbers...that can be expressed (we'll assume
      > two 6N-1 numbers...) as (6a-1)(6b-1) = 6c-1. I've recently noticed that
      > in this case, c can be expressed as: 6ab - b + a. (when there is one 6N+1
      > number, c = 6ab + a - b, and when there are two 6N+1 numbers, c = 6ab + a
      > + b)
      >
      > I'm guessing this thought isn't original and probably several thousand
      > years old...I just don't know where to look.
      >
      > By the way. this expression can be taken further for the multiplication of
      > any two numbers if expressed from 6N-2 to 6N+3.

      Hi Geoff,

      I'm afraid there are so many errors in your algebra that it's hard to know
      where to start...

      It's always a good idea to double-check before posting, by putting in at
      least one set of values.
      Let's take the primes 5 and 11, so a=1, b=2.
      The product is 55.
      Is this of the form 6c-1, with c=6ab - b + a?
      I don't think so !

      Anyway, a formula of that kind can't be right, as interchanging a and b
      gives a different answer, but multiplication is commutative.

      Another general point: I should have thought that whether or not the numbers
      are /prime/ must be irrelevant in equations of the kind you are considering,
      no?
      Try some non-prime numbers.

      Best regards - and better luck next time !

      Mike




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    • mikeoakes2
      ... It s not complicated at all, Geoff, it s just high-school algebra. Let the 2 numbers be x=6*a+a1 and y=6*b+b1, where a1 and b1 are each either +1 or -1.
      Message 2 of 7 , Jun 4 11:27 PM
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        --- In primenumbers@yahoogroups.com, "G Fischer" <golfbum71@...> wrote:
        >
        > Hit send again and realized I misspoke below for the 6N+1 multiplied by 6N-1
        > numbers...
        > c = 6ab + a - b
        > This formula will show differently since a and b can be different... when
        > the numbers are interchanged.
        >
        > As an example: 59*67 = 3953
        > The formula for c in this case is: c = 6ab + a - b
        > a = 10
        > b = 11
        > c = 659
        > c = 6*10*11 - 10 + 11
        > c = 660 + 1
        > c = 661
        >
        > and flopped:
        > a = 11, b = 10, c = 659
        > c = 6*10*11 - 11 + 10
        > c = 660 - 1
        > c = 659
        >
        > Perhaps a rule for whichever number is 6N-1 is added and 6N+1 is subtracted?

        It's not complicated at all, Geoff, it's just high-school algebra.

        Let the 2 numbers be x=6*a+a1 and y=6*b+b1, where a1 and b1 are each either +1 or -1.
        Then x*y = (6*a+a1)*(6*b+b1) = 6*c+c1,
        where c = 6*a*b+a1+b1,
        and c1 = a1*b1 (= +1 or -1).

        Mike
      • djbroadhurst
        ... Not at my high school, where c = 6*a*b + a*b1 + b*a1 David
        Message 3 of 7 , Jun 5 2:41 AM
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          --- 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 4 of 7 , Jun 19 8:30 PM
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            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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