# 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 ... ?