## Re: Multiplying Primes and Numbers in General

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• ... 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 1 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
• ... Not at my high school, where c = 6*a*b + a*b1 + b*a1 David
Message 2 of 7 , Jun 5 2:41 AM
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"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
• 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 3 of 7 , Jun 19 8:30 PM
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Yes.

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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