If three prime numbers, all greater than 3, are in A.P. , then their common difference
(A) must be divisible by 2 but not necessarily by 3;
(B) must be divisible by 3 but not necessarily by 2;
(C) must be divisible by both 2 and 3;
(D) need not be divisible by any of 2 and 3;
Discussion:
Say p, p+d and p+2d are the three primes in A.P. with d as the common difference.
Since p>3, hence p is odd.
If d is odd, then p+d is even. But then p+d cannot be a prime any more. Hence contradiction. Thus d cannot be odd. Hence 2 divides d.
Again p> 3 implies p is not divisible by 3. Hence p is either 1 or 2 modulo 3. Now we wish to check if d is divisible by 3 or not. Suppose it is not then d is also either 1 or 2 modulo 3.
Case 1: p is 1 mod 3 and d is 1 mod 3, then p+2d is 0 mod 3 hence contradiction (as p+2d is a prime greater than 3).
Case 2: p is 1 mod 3 and d is 2 mod 3, then p+d is 0 mod 3 hence contradiction (as p+d is a prime greater than 3)
Case 1: p is 2 mod 3 and d is 1 mod 3, then p+d is 0 mod 3 hence contradiction (as p+d is a prime greater than 3).
Case 2: p is 2 mod 3 and d is 2 mod 3, then p+2d is 0 mod 3 hence contradiction (as p+2d is a prime greater than 3)
Hence d must be 0 modulo 3.
Therefore common difference must be divisible by both 2 and 3.
Answer: (C)
