jasen Posted June 27, 2016 Report Share Posted June 27, 2016 Put number 1 to 21 to the circles, so sum of the numbers in circles at each line are equal. Quote Link to comment Share on other sites More sharing options...

0 Logophobic Posted June 27, 2016 Report Share Posted June 27, 2016 Simple: Spoiler The center 3x3 is any normal magic square of order three. (The numbers 1 to 9, with sum 15 on each row/column) The outer circles are the numbers 10 to 21 in pairs with sum 31. (10+21 ... 15+16) The sum of the numbers in each line is then 46. Alternately: Spoiler The center 3x3 is a magic square using the numbers 7 to 15 with row/column sum of 33, with the remaining numbers in the outer circles having pairwise sum of 22. The sum of the numbers in each line is then 55. Or the center 3x3 is a magic square using the numbers 13 to 21 with row/column sum of 51, with the remaining numbers in the outer circles having pairwise sum of 13. The sum of the numbers in each line is then 64. Question: Spoiler Is there a solution such that the center 3x3 is not a magic square? Quote Link to comment Share on other sites More sharing options...

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## jasen

Put number 1 to 21 to the circles, so sum of the numbers in circles at each line are equal.

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