Source: Original, based on a puzzle in Michael Holt's Math Puzzles and Games, Vol. 1, Page 43.
Disregarding rotations and reflections, there is only one solution:
2 1 4 3 5 7 6 9 8
Hmmm...melikes those: the kind you can do on the corner of a Ronald MacDonald's napkin! Fill an XandO design: A B C D E F G H I Since A+I = B+H = C+G = D+F Then 4(A + I) = 45 - E ; A + I = (45 - E) / 4 Also, 2E = A + I (by combining other equalities). Now, after finishing your Big Mac, you have the possibilities: E A+I 1 11 5 10 9 9 So combos are: 9,1 8,2 7,3 6,4 (since 2E = A + I) Draw a couple more XandO designs, fiddle around to get, around the middle of your 99 cents caramel sundae: 2 1 4 3 5 7 6 9 8 which is the ONLY solution; but which may be upsidedowned and bedroomceilingmirrored into 7 other "appearances". Similarly (but impossible on a napkin!), you can do this with a 5by5, numbers 1 to 25: A B C D E F G H I J K L M N O (hey: we have M in Middle!) P Q R S T U V W X Y ...(A+B-C+D+E) = (F+G-H+I+J) = and so on... Don't feel like typing too much; arranging/combining equalities will lead to: 4M = A + G + S + Y; so we get a M = 13, and all 12 combos = 39. Here's one solution: 25 6 17 21 4 (25+6-17+21+4 = 39) 7 1 9 22 18 15 11 13 14 12 19 23 10 2 5 3 20 16 8 24