3 ms·

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

by pcmaffey 8y ago

anyfoo 8y ago
... I think that statement is only true for the prime number 5. ;)
- pcmaffey 8y ago
  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.
  - dragonwriter 8y ago
    > 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.
    - pcmaffey 8y ago
      > 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 https://en.wikipedia.org/wiki/Senary I expressed it poorly, but not as you state, incorrectly.
      - dragonwriter 8y ago
        > 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.
      - spenczar5 8y ago
        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.
dragonwriter 8y ago
> 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.

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

anyfoo 8y ago

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

pcmaffey 8y ago
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.
- dragonwriter 8y ago
  > 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.
  - pcmaffey 8y ago
    > 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 https://en.wikipedia.org/wiki/Senary I expressed it poorly, but not as you state, incorrectly.
    - dragonwriter 8y ago
      > 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.
    - spenczar5 8y ago
      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.

dragonwriter 8y ago

> 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.