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)**

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