Its explained via modular arithmetic which is a standard tool and area of research in number theory.
Generically if x = Np + q where N and p are integers then it is said that x = q mod p, ie x is q more than a multiple of p. For example 7 = 2 mod 5, since 7 = 5*1 + 2, 23 = 7 mod 16 or 23 = 7 mod 8. It easily explains things like why a number which is divisible by 3 also have the sum of its digits divisible by 3.
We use base 10, which is 1 more than a multiple of 3, ie 10 = 1 mod 3. Now consider a number with 2 digits (ie its between 10 and 99 inclusive), which we'll write as XY (for example if we consider 56 then X = 5, Y = 6 or for 93 X = 9 and Y = 3 etc). It follows that XY = 10*X + Y, so for 56 we have 56 = 5*10 = 6 etc. Now consider this in mod 3, where we have 10 = 1 mod 3. Thus we have XY = X*10 + Y = X*(3*3+1) + Y and taking this mod 3 gives XY = X*1 + Y mod 3 = X+Y mod 3. Thus the 2 digit number XY is divisible by 3 if X+Y is. This works for any number of digits because $$10^{N} = 1$$ mod 3 for all whole numbers N.
For instance, consider 5541. Is this divisible by 3? Well 5541 = 5+5+4+1 mod 3 = 15 mod 3. So 5541 is divisible by 3 if 15 is. Obviously it is but we can use the same method again to see it, ie 15 = 1+5 mod 3 = 6 mod 3 and 6 is divisible by 3, so 15 is, so 5541 is. Its clear that you can do this for any sized initial number, repeating it again and again until you get a single digit number which you can easily determine whether its a multiple of 3.
For instance, is 536392004240671 divisible by 3? We add up the digits, 5+3+6+3+9+2+0+0+4+2+4+0+6+7+1 = 52. Now add its digits up, 5+2 = 7. 7 isn't divisible by 3 and thus neither is 536392004240671. Didn't even need a calculator!
Precisely the same applies to mod 9 arithmetic. If you use hexidecimal (base 16) then it works for mod 5, mod 3 and mod 15. If you use base 12 then it works for mod 11 arithmetic.
So yeah all good Great Wizard of OZ land, but the scare crow still has no brain. If it is given in the prime number lines of the 6n form in mod 3 all prime numbers fall in the 2,8,5 or the 1,7,4 then we could order up all the numbers in the line by pairs of 5,7 and 2,4 and 1,8 . given by the mod of 3 then all multiplication of all pairs would be congruent with 8 which would fall on the 2,8,5 line . given that all sq of prime fall on the 1,7,4 line then it is easy to assume that the pairing in this method would be a pregenerator to sq. interruptions of the prime number field, falling half way between the 2 sq. of the pairing, so like if we take 11 and 13 for example and multiply we would get 143 and in the line of 1,7,4 the sq. of 11 and the sq. of 13 would fall equal distance down the line from 143 and up the line from 143 there for my pea brain can imagine the pattern of prime in a step wise fashion , like building stairs so to speak, or would this just be a fancy way of doing the sieve of Eratosthenes?