Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

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

 Messages: [ Previous | Next ]
 Scott Berg Posts: 2,037 Registered: 12/12/04
The first new theorem Primes
Posted: Sep 18, 2013 11:48 AM

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

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.

Date Subject Author
9/18/13 Scott Berg
9/18/13 Brian Q. Hutchings
9/18/13 Don Redmond
9/19/13 William Elliot