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That would be 5+4 =9

My bad, added 4 to 7.

Yep.

To Ollie's original question - no, they can't all be prime, as exactly one of {a,a+2,a+4} must be divisible by 3.

To see this, do you understand modular arithmetic?

If a ≡ 0 (mod 3), then it's divisible by 3 and hence not prime.

If a ≡ 1 (mod 3), then (a + 2) ≡ 0 (mod 3) and hence is not prime

If a ≡ 2 (mod 3), then (a + 1) ≡ 0 (mod 3) and hence is not prime

If the above doesn't make sense, let me know.

I'm hopng that we're not just doing your homework for you?

Edited by:"Illusionary" 2nd Apri will leave it there for my stupidity

in my head it went 17 19 23 haha

Edited by:"pinkleponkle" 2nd AprThen a + 4 = 21 = 3 × 7

Edited by:"Illusionary" 2nd AprIf you add 2 it will probably make it divisible by 3 and if not the it will be divisible by 5.

A=7 9 11 9/3

A=11 13 15 15/5

A=9 11 13 9/3

A= 13 15 17 15/3 or 5

All primes can’t be even after 2 and cannot end with 5 or 0.

Divisibility by 5 is a red herring, but considering divisibility by 3 is indeed the way to go. See my post above for a full explanation.

Therefore if A+2 is odd, A, A+2 and A+4 are 3 consecutive odd numbers. Out of 3 consecutive odd numbers one will always be divisible by 3.

Where A>3 it eliminates 1,3,5 and 3,5,7 which are the only 2 exceptions.

Yep - just make sure that you understand how to prove your assertion about three consutive odd numbers.