Mathematical puzzle

Do you mean: are they all either one more or else one less than some multiple of 6?

Yes, they are.

How did you arrive at that extraordinary conclusion (and what of 7,11 and 13)?

I'll get to the 7,11,13 problem in a minute...

First things first...

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Claim:
Every prime number greater than 5 is either one less or one more than some multiple of 6.


Proof:
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Consider any prime number "n" which is greater than 5.
Let "r" be the remainder when you divide n by 6.
Let "q" be the whole-number quotient when you divide n by 6.

Note that r is one of the following numbers: 0,1,2,3,4 or 5.

Well, r can't be even, because then n would also be even and could not be prime.
(r even => n=r+6q even)

Also, r can't be 3, because then n would be divisible by 3.

That means r=1 or 5.

If r=1, then n=6q+1 is one more than a multiple of 6.
If r=5, then n=6(q+1)-1 is one less than a multiple of 6.

If you divide "713,713" by 7, then by 11, then by 13, you will get "713" as your answer.

If you divide "abc,abc" by 7, then by 11, then by 13 (where a,b and c are arbitrary digits) you will get "abc" as your answer.
That's because "abc,abc" = 1,001 * "abc". And 1001 = 7 * 11 * 13.