Latest Microsoft Aptitude Question SOLUTION: Pairs of primes separated by a single number are called prime pairs. Examples are 17 and 19. Prove that the number between a prime pair is always divisible by 6 (assuming both numb

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Pairs of primes separated by a single number are called prime pairs.
Examples are 17 and 19.
Prove that the number between a prime pair is always divisible by 6 (assuming both numbers in the pair are greater than 6). Now prove that there are no prime triples.

Asked In Microsoft Naveen (14 years ago)

Solved Tijo Jose Read Solution (3)

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Tijo Jose  (14 years ago)

To prove there are no prime triples...
Let there be a prime triple, say p, p+2, p+4
So, all these are prime by assumption.

Exactly one number is divisible by n from n consecutive counting numbers.
For three consecutive counting numbers exactly one of them is divisible by three.

So, either p or p+1 or p+2 is divisible by 3.
If p or p+2 is divisible by 3, they are not prime.
Or if p+1 is divisible be 3, so is (p+1)+3 = p+4.

So, exactly one on p, p+2, p+4 is divisible by 3. Hence atleast on of them is not prime. So, prime triplets cannot exist.

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Tijo Jose  (14 years ago)

Proof for number between prime pair is always divisible by 6

Let the prime pairs be p and p+2
Obviously none of them is divisible by 3, as they are greater than 6 and are prime. Sine exactly one among three consecutive counting numbers have to be divisible by 3, from among p, p+1, and p+2, p+1 must be divisible by 3. Also, p and p+2 are odd. Hence p+2 is even. All even numbers divisible by 3 are also divisible by 6.

So, p+1 is divisible by 6.

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Tijo Jose  (14 years ago)

Proof of number between prime pair is divisible by 6: Let pair be p and p+2. Since p and p+2 are not divisible by 3 (as they are prime), p+1 is divisible by 3 (by Pigeonhole principle) and also is even (as p and p+2 being prime are odd). All even numbers divisible by 3 are divisible by 6. Hence proved.
Proof that there are no prime triplets: This is proved by contradiction. Let there be a prime triple p, p+2 and p+4. By pigeonhole principle, exactly one of p+1 (and hence p+4), p+2 and p+3 (and hence p) are divisible by 3. But this is not possible as p, p+2 and p+4 were assumed to be prime. H

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