Problem for a six-year-old...

Chris

This is a problem which was given to my six-year-old cousin:
(you'll probably need a bit of paper)

Imagine three circles A, B and C in a straight line. A overlaps B and B overlaps C such that there are 5 distinct regions. You have to place the numbers 1-5, using each number only once, into the five regions, such that the total in each circle is the same. My cousin had to do this for 3 circles and 4 circles (i.e. 7 regions and numbers 1-7)

3 circles is easy... the regions would be 5 1 3 2 4
the total in each circle is six

4 circles is a bit harder and this is when my cousin's mum asked me (I don't think my cousin even understood the problem in the first place ). I got it pretty quickly with trial and error and a little bit of common sense:

7 3 2 5 1 4 6
the total in each circle is 10

So my cousin got his homework right. I suspect it was supposed to just practise their adding, but by now I was quite intruiged. I wondered whether there is a solution for all possible numbers of circles > 2. (2 circles has no solution, and 1 has a pretty obvious solution). By trial and error I worked out the total for 5 circles, which is 11:

9 2 5 4 6 1 7 3 8

By writing a rudimentary computer program, my friend worked out that the total for 6 circles was 14 (if I remember correctly), but I still cannot see a pattern. For 6 circles the numbers in the outermost regions are not the largest two numbers as they have been in the solutions above. None of the rest of my maths set (I'm seventeen, and doing maths/further maths), or my maths teacher, could find a pattern allowing us to predict the arrangement of numbers, or even the total, given the number of circles.

So:
Can it be proven that there is a solution for all numbers of circles > 2?
Can the arrangement and total in each circle be worked out given the number of circles?
Are there ever multiple solutions for a given number of circles (apart from the reflection)?

I, and my maths set, would be very grateful and interested by any insight anyone can provide.
Chris

chronon

For 4 circles 7 2 6 1 3 5 4 is also a solution, with each circle totalling 9.

Chris

Well spotted chronon. Initially I hadn't considered solutions where the largest numbers aren't in the outermost circles, but that's clearly not the best approach.
That answers the question about multiple solutions!

Chris

Chris

Bumpety-bump

trancefishy

this is intriguing. i have putzed around a bit solving this with linear equations, but so far haven't had any success other than a small amount of info pertaining to which regions should equal other possible entries.

i am sure that a rudimentary course in combinatorics would shed a great deal of light on this problem.