Six distinct positive integers \(a,b,c,d,e,f\) are given. Jack and Jill calculated the sums of each pair of these numbers. Jack claims that he has \(10\) prime numbers while Jill claims that she has \(9\) prime numbers among the sums. Who has the correct claim?

(Adapted from a past year Singapore Mathematical Olympiad question)

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## Comments

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TopNewestJack has the wrong claim.

Since the numbers are distinct positive integers, the sum of a pair\(\ge 1+2=3\), hence the primes must be odd. On the other hand, we can't obtain \(10\) odd sums because if \(n\) numbers are odd(\(0\le n\le 6\)), then the number of odd sum is \(n*(6-n)\)(we choose an odd then an even), which doesn't equal to \(10\) for all \(n\), but \(9\) primes is obtainable when \(n=3\).

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Can you find a possible set of six positive integers that satisfies the conditions then?

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yes it suffices to find 3 sets of three primes that are spaced the same way, so take \(1,3,7,4,16,40\).

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