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Topic: The first new theorem Primes
Replies: 3   Last Post: Sep 19, 2013 4:11 AM

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Don Redmond

Posts: 57
Registered: 5/5/11
Re: The first new theorem Primes
Posted: Sep 18, 2013 7:31 PM
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On Wednesday, September 18, 2013 10:48:15 AM UTC-5, Brüder des Schattens Söhne des Lichts wrote:
> The first new theorem is as follows:
>
>
>
> Theorem 1: For any given odd integer that does not contain the integer 5
>
> as a factor, there exists an infinite number of integers (consisting only of
>
> digit 9's, e.g., 999999.) which contain this given odd integer as a factor.
>
>
>
> The informal proof of this theorem is the following:
>
>
>
> Let's designate "B" as any odd integer that does not contain the integer 5
>
> as a factor , and "A" as a slightly larger integer. Furthermore we will
>
> require that A and B have no prime factor in common. If we divide A by B,
>
> we will obtain a quotient consisting of a integer part and a decimal part.
>
> We will designate the integer part of the quotient as "Q" and the decimal
>
> part of the quotient as "D". We will then multiply the decimal part of the
>
> quotient by B, in order to obtain an integer remainder, which we will
>
> designate as "R".
>
>
>
> A/B = Q + D and
>
>
>
> D(B) = R
>
>
>
> Another way to find the value of the integer remainder is to subtract Q(B)
>
> from A
>
>
>
> Eqtn. (1) A1- Q1(B)= R1
>
>
>
> D will always be a repeating decimal. If we multiply our original A by
>
> ten, raised to an integer power equal to the number of digits in one
>
> repetition cycle of D, (designate this new value as A2) and divide A2by B,
>
> we will obtain a new D (call it D2), and a new R (call it R2), which are
>
> the same values as our original R and D respectively. Of course, we will
>
> have a different value for Q (call it Q2).
>
>
>
> A2/B = Q2 + D2 and
>
>
>
> D2(B) = R2
>
>
>
> Another way to find the value of the integer remainder is to subtract Q(B)
>
> from A
>
>
>
> Eqtn. (2) A2- Q2(B)= R2
>
>
>
> At this point, I think that the introduction of a numerical example will
>
> clarify this presentation.
>
>
>
> Let B = (37)(11)(3) = 1221
>
>
>
> Let A1 = (19)3(2) = 13718
>
>
>
> Then A1/B = 13718/1221 = 11.235053235053235053235053235053.
>
>
>
> Q1 = 11
>
>
>
> D1= .235053235053235053235053235053.
>
>
>
> Multiplying B by D1, we obtain R1
>
>
>
> (1221)(.235053235053235053235053235053.) = 287
>
>
>
> R1= 287
>
>
>
> Another way to find the value of the integer remainder is to subtract Q(B)
>
> from A
>
>
>
> Eqtn. (1) A1- Q1(B)= R1
>
>
>
> 13718 -11(1221) = 287
>
>
>
> The number of digits in each repetition cycle of D is 6. Now, if we multiply
>
> A1 by (10)6, we obtain.
>
>
>
> A2 = 13718000000
>
>
>
> Dividing A2by B, we obtain.
>
>
>
> 13718000000/1221 = 11235053.235053235053235053235053.
>
>
>
> Q2 = 11235053
>
>
>
> and
>
>
>
> D2 = D1 = .235053235053235053235053.
>
>
>
> Multiplying B by D2, we obtain the same value for R as in the first step.
>
>
>
> (1221)(.235053235053235053235053235053.) = 287
>
>
>
> R1 = R2= 287
>
>
>
> Another way to find the value of the integer remainder is to subtract Q2(B)
>
> from A2
>
>
>
> Eqtn (2) A2 - Q2(B)= R2
>
>
>
> 13718000000 - 11235053(1221) = 13718000000 - 13717999713 = 287
>
>
>
> Now, we will return to the literal presentation.
>
>
>
> Looking at the second method of obtaining the integer remainder: combining
>
> Eqtn. (1) with Eqtn (2), we can write.
>
>
>
> A1 - Q1(B)= A2 - Q2(B)= R1 = R2 = R
>
>
>
> Or
>
>
>
> A1 - Q1(B)= A2 - Q2(B)
>
>
>
> Rearranging terms
>
>
>
> Q2(B) - Q1(B) = A2 - A1
>
>
>
> Let X represent the number of digits in a repetition cycle of D
>
>
>
> Since we know that A2 = (10)X A1
>
>
>
> then A2 - A1 = (10)XA1 - A1
>
>
>
> and thus A2 - A1 = [(10)X - 1] A1
>
>
>
> and Q2(B) - Q1(B) = [(10)X - 1] A1
>
>
>
> Dividing through by B, we obtain.
>
>
>
> Eqtn. (3) Q2 - Q1 = [(10)X - 1] A1/B
>
>
>
> We have an integer value on the left hand side of Eqtn. (3), and thus we
>
> should also have an integer value on the right hand side of Eqtn. (3). In
>
> order for this to be true, since A and B have no prime factor in common,
>
> then [(10)X - 1] must be evenly divisible by B . [(10)X - 1] will
>
> always be an integer consisting only of digit 9's (e.g., 999999.)
>
>
>
> Back to our numerical example.
>
>
>
> Since A2 = 1000000A1
>
>
>
> A2- A1 = 999999A1
>
>
>
> Then
>
>
>
> Q2(B) - Q1(B) = 999999A1
>
>
>
> Dividing through by B, we obtain.
>
>
>
> Eqtn. (3a) Q2 - Q1 = ( 999999A1)/B
>
>
>
> 11235053 - 11 = (999999)(13718)/1221
>
>
>
> 11235042 = 11235042
>
>
>
> We have an integer value on the left hand side of Eqtn. (3a), and thus we
>
> should also have an integer value on the right hand side of Eqtn. (3a). In
>
> order for this to be true, since A and B have no prime factor in common,
>
> then 999999 must be evenly divisible by B .
>
>
>
> Checking this out in the numerical example, we find.
>
>
>
> 999999/1221 = 819
>
>
>
> An integer quotient value is produced.
>
>
>
> Thus we have shown that our randomly chosen odd integer, 1221, is a factor
>
> of 999999.


If I understand the statement of your result correctly, then this is a theorem
of Crelle. His version allows for concatenated blocks of digits, that is things like 321321321....

Don




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