Remove the all the cards A,2,3,4,5,6,7,8,9 from a deck of cards. There are four of each value for a total of 36 cards. Can you place the cards into a 6x6 grid such that each row, column, and main diagonal has the same sum, and no card value is repeated in any sum. (Let Ace be a value of 1.)
Perhaps this is too unrestrictive? Can it be solved if we also require that each marked 3x2 rectangle also have the same sum?
A68 249 924 735 573 68A 68A 924 249 573 735 A68 ================================================== A) The magic sum must be 30. This is immediate. [KD: (Sum of all cards)/6 = 30] B) The strings with 6 distinct elements and with sum 30 are the permutations of one of the 8 strings, here down denoted by letters from "a" to "i"; the easy proof can be shortened by looking at their complements (strings of length 3 with sum 15): a = 234678, say all but 159 b = 234579, say all but 168 c = A35678, say all but 249 d = A34679, say all but 258 e = A34589, say all but 267 f = A25679, say all but 348 g = A24689, say all but 357 h = A23789, say all but 456 C) The global restriction on the board (exactly 4 cards of each type) can be accomplished only in 3 ways: - C1) the 6 rows are, in some order, permutations of the strings a,b,c,e,f,g; - C2) the 6 rows are, in some order, permutations of the strings a,a,e,e,f,f; - C3) the 6 rows are, in some order, permutations of the strings b,b,c,c,g,g. In particular the strings d and h can appear only as diagonals. Here again the proof simplifies by looking at the complements (they must give twice each number). D) If we look for a solution of type (C2), then columns must be of type (C3); the restrictions killing a lot of permutations, the number of possibilities becomes very low; and we end up with the solution: h f a e f a e d \ | | | | | | / \ | | | | | | / g-- A 6 8 2 4 9 --g b-- 9 2 4 7 3 5 --b c-- 5 7 3 6 8 A --c g-- 6 8 A 9 2 4 --g b-- 2 4 9 5 7 3 --b c-- 7 3 5 a 6 8 --c / | | | | | | \ / | | | | | | \ d f a e f a e h