There are three prime numbers, p, q and r. Further p and q are consecutive prime numbers. In other words, there is no prime number k such that p < k < q. We are also given that p + q = 2r. Find the smallest such triplet (p, q, r).
This is an interesting piece of question related to prime numbers, asked to me by my colleague. The question is as follows:
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8 comments:
Such a distinct triplet is not possible, because: if r is the average of pa dn q, then r is an element of the set that is bounded by p and q; that is, r is greater than p and less than q. But sine that is not possible is this question (no element k : p < k < q), the only possibility is that p=q=r.
Perfect reasoning.
Your question itself is an absurd one. If you add any two prime numbers you will get an even number.(in your case 2r=even). Then r is either odd or even. If r is odd certainly it is divisible by a number. Hence r is not prime. That is why I said your question is absurd.
Your question itself is an absurd one. If you add any two prime numbers you will get an even number.(in your case 2r=even). Then r is either odd or even. If r is odd certainly it is divisible by a number. Hence r is not prime. That is why I said your question is absurd.
Well that the answer friend :)
@2nd Anonymous and Rahul: Every Prime numbers are odd and odd numbers are those which are not divisible by 2 and not that it is divisible by some other number. So reasoing of 2nd anonymous is wrong.
@ravikg, I think 2nd anonymous wanted to say even when he said odd.
@anonymous, well in mathematics there are questions which are called as "paradox" and this question is just one of them. So the answer is indeed that such numbers are not possible :).
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