1 2 3 4

5 6 7 8

9 10 11 12

13 14 15 16

Pick a number in the table. Cross out the numbers in the same row and column as the number you selected. There should be nine numbers left. From those nine numbers, repeat the process three more times so that every number will eventually be crossed out. What is the sum of the four numbers you selected? Will this result always occur? Why or why not?

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## About benvitalis

math grad - Interest: Number theory

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Thanks. I’m glad you enjoyed my posts

The answer is always the same. It is obvious that I can always take one number in each column and one number in each row. Assume that the sum of the numbers is S. Denote f(i) as the number of column of the number chosen in the row i. In your example, if we chose the numbers 4,5,11,14, then f(1) = 4, f(2) = 1, f(3) = 3, f(4) = 2, and A(1,f(1)) + A(2,f(2)) + A(3,f(3)) + A(4,f(4)) = 34.

Note that A(a,f(a)) + A(c,f(c)) = A(a, f(c)) + A(c, f(a)). The statement means that swapping the column between any two rows does not change the sum, i.e., if I chose 4 in the first row and 5 in the second, I can swap them and chose 1 for the first row and 8 for the second, and the sum does not change.

Therefore, with enough swapping, S = the number on the diamater: S = 1+6+11+15 = 34.

The problem can be generalized to nxn matrix.