# A partition of 1000 into six parts with least and greatest product possible

Find six positive natural numbers, not necessarily distinct, whose sum is 1000 and which, if placed appropriately on the vertices of the following 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).

Find the solutions in which the product of the six numbers is as small and large as possible.

• what's an edge? how many does that graph contain? – Jasen Mar 29 '18 at 20:02
• @Jasen: Lines (in this case straight) joining circles, better known as vertices in graph theory terminology. – Bernardo Recamán Santos Mar 29 '18 at 20:28
• should I be seeing two or or five edges? some graph theory allows more than two nodes on an egde. – Jasen Mar 29 '18 at 20:33
• You rejected my (now deleted) answer with the reason that "vertices which have no edge between them have no common divisor, hence no more than two of them can be even." Where is that said exactly in your puzzle? If so, how would two numbers be equal ("not necessarily distinct")? Sorry if I'm missing something obvious to you, but I'm not used to mathematical jargon and I might not be alone. – xhienne Mar 29 '18 at 20:43
• Am I right to think that the three numbers in the top, left and right circles are interchangeable? They are all in degree 1 vertices and all connect to the central circle of the cross. – nickgard Mar 29 '18 at 20:55

I wrote a program and it produced:

The products are:

$53,361,000$ and $20,091,608,390,700$.

To get a small product we want one large number and lots of small numbers. Conversely, to get a large product we want all the numbers to be close to $1000/6=166.66\dots$ I used this to reduce the search space. Since it wasn't a full search it's always possible I missed better solutions.
• Through computer search I can confirm @nickgard's results. The same minimum can be also achieved with $924,50,11,7,5,3$. – Freddy Barrera Mar 30 '18 at 13:11