## Multiplying Primes and Numbers in General

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• 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
Message 1 of 7 , Jun 3, 2013
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.

Hopinh to get smarter on this stuff.

Thanks for any help.

Cheers,
Geoff
• ... 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
Message 2 of 7 , Jun 4, 2013
--- 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
• 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
Message 3 of 7 , Jun 4, 2013
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...

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.

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.

Cheers,
Geoff

-----Original Message-----
From: mikeoakes2
Sent: Tuesday, June 04, 2013 12:04 AM
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

------------------------------------

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
The Prime Pages : http://primes.utm.edu/

• 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 4 of 7 , Jun 4, 2013
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
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...

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.

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.

Cheers,
Geoff

-----Original Message-----
From: mikeoakes2
Sent: Tuesday, June 04, 2013 12:04 AM
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

------------------------------------

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
The Prime Pages : http://primes.utm.edu/

------------------------------------

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
The Prime Pages : http://primes.utm.edu/

• ... 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 5 of 7 , Jun 4, 2013
--- 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 6 of 7 , Jun 5, 2013
"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 7 of 7 , Jun 19, 2013
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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