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?
Source: Original.
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