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Old 2nd July 2013, 09:56 PM   #1 (permalink)
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Maximum number of prime sums.

You are given 8 distinct positive integers. You will take 2 different numbers from the 8 and sum them up (i.e. pairwise sums of the 8 distinct numbers), what is the maximum number of sums that will be prime?

Note: The sums need not be distinct, so there can be like two numbers, 2,5 , adding to 7 and 3,4 adding to 7, and these will be considered two different sums.

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Old 3rd July 2013, 01:56 AM   #2 (permalink)
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Re: Maximum number of prime sums.

8 choose 2 is 28. So the obvious upper bound is 28 prime sums. But definitely can go lower because a prime number (ignoring 2) must be an odd number. So we can only get a prime number by adding an odd number and an even number. E.g.

Four odd, four even = 16 possible odd sums
Five odd, three even = 15 possible odd sums
Six odd, two even = 12
Seven odd, one even = 7
All odd = 0 possible odd sums

So a firmer upper bound is 16 possible prime sums.

Don't know if can go lower or not

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Old 3rd July 2013, 09:33 PM   #3 (permalink)
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Re: Maximum number of prime sums.

Yes 16 is the correct answer, nothing more than that, but can you justify that it is the answer?(quite easy to justify though)


Last edited by megaman123; 3rd July 2013 at 09:33 PM.
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Old 3rd July 2013, 10:57 PM   #4 (permalink)
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Re: Maximum number of prime sums.

Wah okay, I thought would go lower. Okay I found these eight numbers:

1 3 7 9
4 10 100 190

Four odd and four even. Add any odd to even, will get a prime. So really can get 16 primes. They are:

5, 7, 11, 13
11, 13, 17, 19
101, 103, 107, 109
191, 193, 197, 199

Not easy to find

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Old 4th July 2013, 08:12 AM   #5 (permalink)
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Re: Maximum number of prime sums.

Here's my solution:
Like what Madhat said: 2 can be ignored as the minimum sum you get is larger or equal to 1+2=3.
So we are looking for odd prime sums, which means that we are looking for even number and an odd number sum.
Suppose there are k odd numbers, so 8-k even numbers. So the number of odd sums is k(8-k) only.
k(8-k)=8k-k^2= -(k-4)^2+16
So, 16 is the most odd sums when there are 4 odd and even numbers.
The set of {1,2,3,4,9,10,58,69} gives 16 odd prime sums.(Or use what Madhat has)
Hence 16 is the maximum prime sums.

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Old 4th July 2013, 03:30 PM   #6 (permalink)
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Re: Maximum number of prime sums.

Oh ya maximize k(8-k) is good. I didn't think of that.

Btw, Megaman, have you ever tried to prove Goldbach's conjecture? I mean, the conjecture that every even number (starting from 4) can be expressed as the sum of two primes. Like:

4 = 2+2
6 = 3+3
8 = 3+5
10 = 3+7 or 5+5
etc

Just wondering whether you tried before. I tried many times but can get nowhere. It's a very interesting problem though because it looks so simple, like should not be too hard to prove. But so far nobody has managed to prove it.

http://en.wikipedia.org/wiki/Goldbach's_conjecture

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