# Base Twelve and the prime Numbers

## by Don Hammond

The dozenal base quickly and easily reveals a fundamental property of prime numbers.

For natural numbers in base twelve greater than 3

• All numbers terminating with even digits are divisible by 2 and so are not prime.
• All odd numbers terminating with 3 or 9 are divisible by 3 and so are not prime.
• There exist prime numbers of two or more digits which terminate with 1, 5, 7 or E.
This means that prime numbers written in base twelve will end in 1, 5, 7 and eleven (but not, of course, that all numbers ending in these digits will be prime).

Hence the set of natural numbers terminating with 1, 5, 7 or E must contain all prime numbers greater than 3, and excludes all odd numbers divisible by 3.
It follows that this is the minimum set to contain all primes greater than 3.

Rearranging the terminal digits as 5, 7 and E, 1 shows the set to be of the form

(6n ± 1)

Therefore the minimum set of natural numbers to contain all primes is:

{ 2, 3, (6n ± 1) } n ∈ N.

This last statement is a factual property of prime numbers and is therefore true regardless of the number-base.
It also explains the occurrence of "twin primes", since the only possible position for primes greater than 3 are "each side" of the multiples of 6.

The fact that prime-number positions are completely controlled by 6 (itself the product of 2 and 3, and the companion of our dozenal base) is often not realized even by those with an interest in the subject.
It is never found in school text-books, and even Hall and Knight do not mention it in their "Higher Algebra", which is regarded as a standard work.

I wonder why ... ?