# A partition of 80 into seven parts

The sum of seven positive integers, not necessarily distinct, is 80. If placed appropriately on the vertices of this graph, two of them will be joined by an edge if and only if they have a common divisor greater than 1 (that is, they are not relatively prime). In non-decreasing order, what are those seven integers? • For those interested, this is were it all began:youtube.com/watch?v=6Giujua_s80&t=18s Dec 1 '18 at 18:29
• Is there any reason to the strange phrasing non-decreasing order? You can just say "increasing order". Dec 1 '18 at 21:15
• @Hugh It's maths-y language... Strictly speaking, the number sequence does not always increase, since the second 14 isn't larger than the first. But it never decreases. Dec 1 '18 at 22:06
• @Angkor ah, yes. I had not yet looked at the solution provided by Dr. Xorile and so I did not see that there is a duplicate number. The choice of words makes a lot more sense now. Dec 1 '18 at 23:12

I don't know if this answer is unique, but it seems to work

1,5,7,9,14,14,30

Here it is in place:

There seems to be some confusion in the other answers, so let me spell it out my thinking a little. OP specified that the numbers would be "joined by an edge if and only if they have a common divisor greater than 1".

In other words, if they are joined, then they have a common divisor greater than 1. And not otherwise.

So:

• 7 and 14 are joined (gcd (greatest common divisor) is 7)
• 7 and 30 are not joined (gcd is 1)
• 7 and 9 are not joined (gcd is 1)
• 7 and 5 are not joined (gcd is 1)
• 14 and 14 are joined (gcd is 14)
• 30 and 14 are joined (gcd is 2)
• 30 and 9 are joined (gcd is 3)
• 14 and 9 are not joined (gcd is 1)
• 1 is not joined to anything (gcd is always 1)
• etc.

In order to solve this, I realized that the number in the middle must have at least 3 distinct prime factors, so that it could share those factors with the different arms and have those arms not share a factor.

2 x 3 x 5 is the smallest way to do this. The challenge was going to be keeping it below 80. So I made the 2 go to the left where more factors would come into play and the 3 and 5 go right. That gets us to here:

Now $$x$$ needs to share a divisor with the leftmost position. It cannot be 2, 3, or 5, so 7 is a reasonable guess:

Now we add that up so far and it comes to 73. Mmm. That would make the final number 7, which would be sharing a factor with some of the numbers on the board already. In fact, any number less than 7 would share a factor with something... except 1. So I changed 3 to 9 ($$3^2$$). This doesn't change any of the divisor logic - numbers will still either be relatively prime or not. But it leaves the final number equal to 1. Hence the solution. Probably unique, looking through this logic.

• Quick! I think it'd be nice to show how they fit on the graph. Could you also explain how you got to this answer? Dec 1 '18 at 17:09
• The answer is unique. Dec 1 '18 at 18:30

Seems like there are a lot of possible solutions - but assuming the two connected numbers common divisor cannot be one of the two numbers, and that you cannot repeat a number, I got this:

      6                 10

4            12         36

8                 14


Every number shares a divisor of 2, which is more than 1.