The numbers 1, 2, ... , $n^{2}$ are arranged in an $n \times n$ array in the following way: Pick n numbers from the array such that any two numbers are in different rows and different columns. then the sum of these numbers are $n \sum k = 1 a_{k j k} = \frac{n (n^{2} + 1)}{2} .$
If true then enter $1$ and if false then enter $0$
1 2 3 ... n
n+1 n+2 n+3 ... 2n
$⋮$ $⋮$
$n^{2} - n + 1$ $n^{2} - n + 2$ $n^{2} - n + 3$ ... $n^{2}$

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Solution

If we denote by $a_{i j}$ the number in the i th row and j th column then $a_{i j} = (i - 1) n + j$ for all i, j = 1, 2, ... , n. Because any two numbers are in different rows and different columns, it follows that from each row and each column exactly one number is chosen. Let $a_{1 j 1}, a_{2 j 2}, . . ., a_{n j n}$ be the chosen numbers, where $j_{1}, j_{2}, . . ., j_{n}$ is a permutation of indices 1, 2, ... , n. We have $n \sum k = 1 a_{k j k} = n \sum k = 1 ((k - 1) n + j k) = n n \sum k = 1 (k - 1) + n \sum k = 1 j k .$ But $n \sum k = 1 j k = \frac{n (n + 1)}{2}$
since $j_{1}, j_{2}, . . ., j_{n}$ is a permutation of indices 1, 2, ... , n. It follows that $n \sum k = 1 a_{k j k} = \frac{n (n^{2} + 1)}{2} .$

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