# Magic square variation

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?

• 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 – Alex.vollenga Feb 7 '18 at 20:20

## 1 Answer

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.