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Consider the squares of an 8 *8 chessboard filled with the numbers 1 to 64 as in the figure below. If we choose 8 squares with the property that there is exactly one from each row and exactly one from each column, and add up the numbers in the chosen squares, show that the sum obtained is always 260.

Note by Sayan Chaudhuri
4 years, 1 month ago

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Turn each number into $$8 \cdot x + y$$ where $$0 \leq y < 8$$. $$x$$ turns out to be the row number, starting from $$0$$, and $$y$$ is the column number, starting from $$1$$. Since each row and column is picked exactly once, then ANY configuration's sum would be $$8 \cdot \displaystyle \sum_{i=0}^7 i + \displaystyle \sum_{i=1}^8 i = 8 \cdot 28 + 36 = 260$$.

cmiiw. · 4 years, 1 month ago

i think u r right,hats off · 4 years, 1 month ago

Nice invariant problem. I think of some variants for this problem. a) instead of $$8\times 8$$, find a formula for an $$n\times n$$ table. b) instead of $$1$$ to $$n^2$$, try to find a formula for $$n^2$$ consecutive integers. c)instead of consecutive integers, try to find a formula for an arithmetic progression with $$n^2$$ terms. :) · 4 years, 1 month ago

the problem turns out 2 B more spiral,for me that would be very difficult 2 solve · 4 years, 1 month ago

its not spiral, its leftmost cell to rightmost cell of each row then going to next row from top to bottom. actually b is the most similar to the original problem. · 4 years, 1 month ago

The only way of choosing the 8 numbers is by taking the diagonals. The sum of numbers on the diagonals is 260. · 4 years, 1 month ago

no, thats wrong · 4 years, 1 month ago