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# Combinatorial Question

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1. 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?

2. (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.
3. 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

4. (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.
5. (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.
6. 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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