I have a rather difficult variation of magic squares: In the below image, all numbers from 1 to 24 must be placed in the 24 closed areas, in such a way that all numbers in areas of each circle must sum to 80. Each number must be placed only once.


Some observations: Let's consider two "opposite" circles. These two do not have any areas in common, therefore these must have 2 distinct subsets of 8 numbers each. Similarly, the outer areas to the left and right of these circles, also have 8 distinct areas so this is the 3rd subset of 8 numbers. Since all numbers from 1 to 24 add up to 300, and the 2 above distinct circles must have a sum of 160, the remaining 8 areas (4+4 to the left and right of the distinct circles) must have a sum of 140.

The numbers at the 6 outmost areas of each of the 6 outer circles, appear only once. The numbers at the 6 areas right below them, appear twice. The areas right below these areas appear 3 times each, and the numbers at the inner thin areas that look like flower petals appear 4 times each. So we have x+2y+3z+4w=7*80 since we have 7 circles. Each of x, y, z and w is a sum of 6 distinct numbers.

Finally, the inner circle consists of 12 distinct areas, which means that we have to use the smaller numbers (to get a sum of 80).

Is there any way to continue?

  • $\begingroup$ I think the inner circle has the numbers 1,2,3,4,5,6,7,8,9,10,11 and 14 that sum up to 80. This leaves the rest for the outer circles $\endgroup$ Feb 7, 2018 at 20:20

1 Answer 1


This puzzle has no solution.

Proof: Using your labels, let x represent the sum of the outermost areas, which are part of only one circle. Then let y be the sum of the areas right below them, z be the sum of the areas below that, and w be sum of the innermost "petal" shapes.

Since z+w must equal 80, between them they contain the numbers 1-10 and either 11,14 or 12,13. Suppose w contains the numbers 1-6 and z contains the rest of that set. Then we have z+w=80 and 3z+4w=261.

If we fill out y with the smallest remaining numbers (whether that means 12,13,15-18 or 11,14,15-18), we will have y=91 and 2y=182. Finally, 19-24 will fill out x, with a sum of 129.

At this point we have x+2y+3z+4w=572. But as you pointed out, since these numbers together make 7 circles, we require the sum to be 560. The only way to reduce the sum is to move smaller numbers from y to z, but doing so would violate the rule that z+w must equal 80.

We have no way to proceed, so the puzzle has no solution.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .