```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  andD(B) = RAnother way to find the value of the integer remainder is to subtract  Q(B) from AEqtn. (1)   A1- Q1(B)= R1D 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  andD2(B) = R2Another way to find the value of the integer remainder is to subtract  Q(B) from AEqtn. (2)   A2- Q2(B)= R2At this point, I think that the introduction of a numerical example will clarify this presentation.Let B = (37)(11)(3) = 1221Let A1 = (19)3(2) = 13718Then A1/B = 13718/1221 = 11.235053235053235053235053235053.Q1 = 11D1= .235053235053235053235053235053.Multiplying B by D1, we obtain R1(1221)(.235053235053235053235053235053.) = 287R1= 287Another way to find the value of the integer remainder is to subtract  Q(B) from AEqtn. (1)   A1- Q1(B)= R113718 -11(1221) = 287The number of digits in each repetition cycle of D is 6. Now, if we multiply A1 by (10)6, we obtain.A2 = 13718000000Dividing A2by B, we obtain.   13718000000/1221 = 11235053.235053235053235053235053.Q2 = 11235053andD2 = D1 = .235053235053235053235053.Multiplying B by D2, we obtain the same value for R as in the first step.(1221)(.235053235053235053235053235053.) = 287R1 = R2=  287Another way to find the value of the integer remainder is to subtract  Q2(B) from A2Eqtn (2)    A2 - Q2(B)= R213718000000 - 11235053(1221) = 13718000000 -  13717999713   =     287Now, 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 = ROrA1 - Q1(B)= A2 - Q2(B)Rearranging termsQ2(B) - Q1(B) = A2 -  A1Let  X represent the number of digits in a repetition cycle of DSince we know that A2 = (10)X  A1then  A2 -  A1 = (10)XA1 -  A1and thus  A2 -  A1 = [(10)X  - 1] A1and  Q2(B) - Q1(B) =  [(10)X  - 1] A1Dividing through by B, we obtain.Eqtn. (3)    Q2 - Q1 = [(10)X  - 1] A1/BWe 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 = 1000000A1A2-  A1 = 999999A1ThenQ2(B) - Q1(B) = 999999A1Dividing through by B, we obtain.Eqtn. (3a)    Q2 - Q1 = ( 999999A1)/B11235053 - 11 = (999999)(13718)/122111235042 = 11235042We 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 =   819An integer quotient  value is produced.Thus we have shown that our randomly chosen odd integer, 1221, is a factor of 999999.
```