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    Good afternoon,

    I have 6 different non-repeated numbers from a pool of 1 to 59 numbers, let's say numbers n1...n6 for example.

    Now knowing that if we add any 2 of the numbers together to equal a 3rd number there are 20 different combinations available.

    These are...

    n1 + n2 = n3
    n1 + n2 = n4
    n1 + n2 = n5
    n1 + n2 = n6
    n1 + n3 = n4
    n1 + n3 = n5
    n1 + n3 = n6
    n1 + n4 = n5
    n1 + n4 = n6
    n1 + n5 = n6
    n2 + n3 = n4
    n2 + n3 = n5
    n2 + n3 = n6
    n2 + n4 = n5
    n2 + n4 = n6
    n2 + n5 = n6
    n3 + n4 = n5
    n3 + n4 = n6
    n3 + n5 = n6
    n4 + n5 = n6

    My question is, what is the maximum count of any 2 seperate numbers being added together to make a 3rd number from the 6 numbers, knowing we can use numbers 1 to 59 please.

    I hope this makes sense?

    Thanks in advance.
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    (Original post by PAB9)
    My question is, what is the maximum count of any 2 seperate numbers being added together to make a 3rd number from the 6 numbers, knowing we can use numbers 1 to 59 please.
    I think it's clear that an upper bound on the number is 10.

    I suspect the final answer is 6, but can't prove it.

    An example is: Let the set be {1,2,3,4,5,6}, in order.

    Then,
    n1+n2=n3
    n1+n3=n4
    n1+n4=n5
    n1+n5=n6
    n2+n3=n5
    n2+n4=n6

    6 in all.
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    Thanks for the reply ghostwalker, it is appreciated.

    I was thinking along the lines of 6 but wondered if there was something I was missing considering that the numbers were from 1 to 59.

    I applied the same logic to 3 numbers added together to equal a 4th number & 4 numbers added together to equal a 5th number.

    I believe for 3 numbers added together there are 4 combinations if the numbers were 1,2,4,7,10,13 for example...

    n1+n2+n3=7
    n1+n2+n4=10
    n1+n2+n5=13
    n2+n3+n4=13

    ...likewise, for 4 numbers added together there are 2 combinations if the numbers were 1,2,3,4,10,16 for example...

    n1+n2+n3+n4=10
    n1+n2+n3+n5=16

    Thanks for your time.
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    (Original post by PAB9)
    Thanks for the reply ghostwalker, it is appreciated.

    I was thinking along the lines of 6 but wondered if there was something I was missing considering that the numbers were from 1 to 59.

    I applied the same logic to 3 numbers added together to equal a 4th number & 4 numbers added together to equal a 5th number.

    I believe for 3 numbers added together there are 4 combinations if the numbers were 1,2,4,7,10,13 for example...

    n1+n2+n3=7
    n1+n2+n4=10
    n1+n2+n5=13
    n2+n3+n4=13
    Also, 1,2,5,8,11,14.

    ...likewise, for 4 numbers added together there are 2 combinations if the numbers were 1,2,3,4,10,16 for example...

    n1+n2+n3+n4=10
    n1+n2+n3+n5=16
    I think it would be fairly easy to show that with 4 numbers, 2 is the maximum.
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    (Original post by PAB9)
    I was thinking along the lines of 6 but wondered if there was something I was missing considering that the numbers were from 1 to 59.
    I can see a way of proving that the max number is 6 now for adding just two numbers, based around showing the first two (lowest) cannot be written as a sum, the third and fourth can only be written in one way as a sum of the lower numbers, and the fifth and sixth can only be written in at most two ways. Hence 6.
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    Thanks ghostwalker for your time and input, it is very much appreciated.
    Enjoy the rest of your long weekend and thanks again.
 
 
 
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