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Consider the squares of an $8 \times 8$ chessboard filled with the numbers $1,2,3,4 \ldots ,64$ in sequential order.

If we choose $8$ squares with the property that there is exactly $1$ from each row and exactly $1$ from each column, and add up the numbers in the chosen squares, show that the sum obtained is always $260$.

Please provide full solution and explanation.

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    $\begingroup$ Without specifying how the squares were numbered, this is false: put $1$ in square $a1$, $2$ in square $b2$, etc. Then picking the diagonals the sum is only $1 + \dots + 8 = 36$. $\endgroup$ – grantfgates May 9 '14 at 13:21
  • $\begingroup$ The numbers are entered in sequential order. $\endgroup$ – Calvin Lin May 9 '14 at 13:23
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If we put the numbers in sequential order, the numbers of the row $i$ are of the form $$8i + j $$ $$i \in \lbrace 0,1,2, \ldots 7\rbrace$$ $$j \in \lbrace 1,2,, \ldots 8\rbrace$$

If we choose $8$ squares with the property that there is exactly one from each row and exactly one from each column, with this notation the total is $$8 \cdot \biggl( \sum_{i=0}^{7}i \biggr) + \sum_{j=1}^{8} j = 260$$

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Hint: work in $\pmod 8$

Hint: $260=28\cdot 8 + 36$

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