Mike Jones commented on my post over on the T-blog:

It’s actually not too difficult to show that all primes will end in 1 or 5 in base 6 (although not all numbers ending in 1 or 5 are primes), with the exception of 2 and 3.

We can categorically eliminate numbers that end in 0, 2, and 4. These are non-odd numbers. No prime other than 2 will be even.

That leaves 1, 3, and 5.

Let Xn be any number that ends with the digit 3 in base 6, except 3. This number may be represented as:

Xn = 3 + 6n, where n > 0

We can factor 3 out of the expression:

Xn = 3(1 + 6n), where n > 0

Therefore, all numbers ending in the digit 3 in base 6 are divisible by 3 => No prime numbers will end in 3 in base 6.

That leaves 1 and 5.

Your hypothesis is fact.

:D

This got me to thinking about my other hypothesis that all P numbers are either prime or have factors that are other P numbers. And now I’ve been able to prove that.

Remember that I previously defined a P number as: “The P class is defined as any number that is N +/- 1″ and N was defined as “a positive number that is divisible evenly by 6.” In reality, this definition is too restrictive, so allow me to redefine it slightly:

A P class number is any number in base-6 whose last digit (read left to right) is either a 1 or a 5.

Because this is a base-6 definition, everything in my previous definition still works (N is a multiple of 6 in base-10, so all N numbers will convert to ending in a 0 in base-6; N + 1 or N - 1 in base-10 will convert to a number that ends in either 1 or 5 in base-6 respectively).

Under this new definition of a P class number, we can also include the number 1 (as it is a number whose last digit is either a 1 or a 5).

Now, my hypothesis is that a P class number must be either prime or its factors must also be P class numbers. Since prime numbers are divisible by 1 and 1 is a P class number, we can actually exclude the “prime” portion from the above and simply say: A P class number can only be divisible by another P class number.

To prove this, we need to demonstrate that A) any P class number multiplied by another P class number will give us yet a third P class number (which I demonstrated in my previous post) and B) no non P class number can ever be multiplied by any other number (P class or not) to form a P class number. So to prove it:

1. Any number that ends in digit x multiplied by a number that ends in 1 will have x as it’s terminating digit. Therefore, if multiplying by a P class number that ends in 1 to get another P class number, you must multiply by a number that ends in either 1 or 5. Therefore, multiplying a P class number that ends in 1 requires multiplying by another P class number to yield a P class number.

2. In an even base system (and base 6 is an even base system since 6 is even), any number x multiplied by any even number will be an even number. Examples (in base-6):

3 x 2 = 10 (x = 3)
4 x 2 = 12 (x = 4)
12 x 4 = 52 (x = 12)

Therefore, any number multiplied by 0, 2, or 4 must be an even number and cannot end in either 1 or 5.

3. An even number multiplied by a number that ends in 3 will end in 0 in base-6 (we don’t really need to demonstrate this step, since even numbers were excluded in step 2, but I’m including it just to be complete); an odd number multiplied by a number that ends in 3 will end in 3 in base-6. Example:

2 x 3 = 10.
3 x 3 = 13.
4 x 3 = 20.
5 x 3 = 23.

Therefore, no number multiplied by a number ending in 3 could end in 1 or 5.

4. This leaves only 5. Taking a look at the multiplication table of 5 in base-6 is quite interesting:

5 x 1 = 5
5 x 2 = 14
5 x 3 = 23
5 x 4 = 32
5 x 5 = 41
5 x 10 = 50

If you note, this has the same type of thing we find in base-10 systems with the number 9. In base 10, you can construct the multiples of 9 by simply writing a column of numbers from 0 - 9. Then, to the right of each number, write the inverse (9 - 0) next to the previous number. Hence:

09
18
27
36
45
54
63
72
81
90

Here, the same thing is seen in base-6, only with the range of 0-5:

05
14
23
32
41
50

Likewise, just as the digits in the 9 sequence add up to 9 (i.e. 18 is 1 + 8 = 9, 36 is 3 + 6 = 9), so in base-6 in the 5 sequence the digits add up to 5 (23 is 2 + 3 = 5; 41 is 4 + 1 = 5). I’m guessing (though I haven’t tested it) that this is the case of any even base system. (I.e., given base N where N is an even number, then the multiplication table of N - 1 will display the above behavior.)

In any case, that’s just something I discovered but it’s not relevant to the current discussion. What is relevent is this: a number that ends in 5 can only end in 1 if multiplied by a number that ends in 5; and a number that ends in 5 can only end in 5 if multiplied by another number that ends in 1. Therefore, the only way to make another P class number multiplying by a P class that ends in 5 is if you multiply it by another P class number.

Conclusion: A P class number can only be divisible by another P class number.