Can we put positive integers $1,2,3, \cdots 64$ into $8 \times 8$ grids such that the sum of the numbers in any $4$ grids that have the form like $T$ ( $3$ on top and $1$ under the middle one on the top, this can be rotate to any direction) can be divided by $5$?
Problem
Source: China North MO
Tags: rotation, combinatorics unsolved, combinatorics
tipe
14.08.2006 08:49
The answer is no. Suposse it is possible: Let's enumerate de grids in the following way: in the first row with 1,2,3,4,5,6,7,8; in the second row with 9,10,11,12,13,14,15,16, etc. Let $A$ the number in the square 2, taking the $T$'s (9,10,11,2) and (9,10,11,18) it follows that the number in the square 18 is congruent to $A$ mod 5, analogous the numbers in the squares 34,50,52,36,20,4,22,6,38,54,56,40,24 are congruent to $A$ mod 5, the trouble is in 8, since the numbers in 7 and 23 are congruent, and taking the $T$'s (6,7,8,15) and (22,23,24,15) and using that the numbers in 6,22,24 are congruent mod 5, it follows that the number in 8 is also congruent the number in 24, i.e. is congruent to $A$. Analogous the numbers in: 1,3,5,7,17,19,21,23,33,35,37,39,49,51,53,55 are conguent mod 5, say to $B$. 9,11,13,15,25,27,29,31,41,43,45,47,57,59,61,63 are congruent mod 5, say to $C$. 10,12,14,16,26,28,30,32,42,44,46,48,58,60,62,64 are congruent mod 5, say to $D$. It's obvious that $A;B;C;D$ are pairwise distinct mod 5 (if not we have 32 numbers in 1,2,3,...,64 mod 5) , so we have only 4 possible residues mod 5, it is a contradiction because for example 50,51,52,53,54 are pairwise distinct mod 5. $Tipe$