Date: Sep 18, 2013 11:48 AM Author: Scott Berg Subject: The first new theorem Primes 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.