Combinatorics #4

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)

#Combinatorics

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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Jack 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 $n=3$ .

Xuming Liang - 6 years, 11 months ago

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

Victor Loh - 6 years, 11 months ago

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

Xuming Liang - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$