## Card Magic Square

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.

I received a very nice solution from Claudio Baiocchi which solves all aspects of the problem.  Instead of just solving for the 2x3 rectangles, he found a solution where every line of three cards has a sum of 15.  This of course meets the requirements for both the 2x3 and the 3x2 rectangles.
```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```

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