It does not seem possible to use consecutive integers to solve this problem. The solution using the smallest set of positive integers is discussed here.

The 4-circle set has 8 intersection points, so the best one could hope for is a solution using {1...8}. There is none, nor is there a solution using numbers taken from {1...9}, but there are quite a few if {1...10} is used. This is a bit hard to explain in ASCII, but the labels are roughly in the following spots:

1 3 6 10 9 8 4 2and the sum of each circle is 25.

Do you think the four-circle problem could be solved with consecutive numbers in which the initial number is not 1? We believe it is impossible.

The key is that two of the circles use 5 points and two use 4. Now suppose it can be solved with the set {M, .... M+7}, and imagine Now suppose it can be solved with the set {M, .... M+7}, and imagine M is "large". Then, by the magic condition, one of the 5-point circles must equal one of the 4-point ones - say

b+c+f+g+h = b+d+e+h

(which is, in fact, one of the equations, if the points of intersection are labeled a through h in a certain way)

Now, write a = M+a', b=M+b', etc., where a', b', etc. are now integers in the range 0 to 7. The equation becomes

5M+(b' + c' + f' + g' + h') = 4M+(b' + d' + e' + h')

or

M = (d' + e') - (c' + f' + g')

where I've deleted the terms that cancel (this isn't important, but just for clarity I did it). But now we see that a', b', etc. are small numbers and so there is no way the expression on the RHS can be equal to the large number M. In this equation, in fact, M only needs to be 11 or larger for it to be impossible, since the largest the RHS can be is (7+6)-(2+1+0) = 10.

This proves that it can't be done with M large. The overall proof is completed by showing, by computer, that it's also impossible for small M, up to a limit where the above argument kicks in. I just did that also, so I think we have Q.E.D.

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