Hacker News new | past | comments | ask | show | jobs | submit login

Because primes exist on a base 6 numeric scale. Where all prime numbers are multiples of 1 and 5.



> Because primes exist on a base 6 numeric scale. Where all prime numbers are multiples of 1 and 5.

What you seem to mean to claim is that primes > 3 are all congruent to either 1 or 5 modulo 6.


... I think that statement is only true for the prime number 5. ;)


Base 6:

1,2,3,4,5,6

7,8,9,10,11,12

13,14,15,16,17,18

19,20,21,22,23,24

25,26,27,28,29,30

See a pattern? 2 & 3 are the only prime numbers that are an exception. All others are multiples of 1 and 5.


> All others are multiples of 1 and 5.

Assuming you used “and” when you meant “or”, that's trivially true (and redundant) in that all numbers (irrespective of base, which has no effect on this) are integer multiples of 1.

But no primes other than 5 are integer multiples of 5, in any base.


> Senary may be considered interesting in the study of prime numbers, since all primes other than 2 and 3, when expressed in senary, have 1 or 5 as the final digit.

https://en.wikipedia.org/wiki/Senary

I expressed it poorly, but not as you state, incorrectly.


> I expressed it poorly, but not as you state, incorrectly.

No, really, it is completely incorrect to use “multiple of X in base Y” to mean “have X as the final digit in base Y” (which is equivalent to “is congruent to X modulo Y.”)

13 is not, in base 10 (or anywhere else), a multiple of 3.[0]

That's just not what “multiple” means.

[0] Well, the number denoted by the digits “13” in any base that is itself a multiple of 3—other than base 3 itself where “13” is not a valid number—is a multiple of 3, obviously, but we're talking about the number represented by “13” in base 10.


Having x as a final digit does not make a number a multiple of x. For example, 13 is not a multiple of 3 in base 10.




Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact

Search: