Let $A$ be the $5\times 5$ matrix
$\begin{bmatrix}11& 17 & 25 & 19 & 16\\ 24& 10 & 13 & 15 &3 \\ 12& 5 & 14 & 2 & 18\\ 23 & 4 & 1 & 8 & 22\\ 6& 20 & 7 & 21 & 9\end{bmatrix}.$ Find the group of $5$ elements, one from each row and column, whose minimum is maximized and prove it so.
I think this problem involves maximizing the choices picked per column. For the $2$nd and $4$th columns, we may only pick one of $20$ and $21.$ Similarly, for the $2$ and $4$th rows, we may only pick one of $23$ and $24.$ For the first row, we can pick only one of $25, 16, 17, 19.$ how can i continue from here?
The problem is asking for the smallest number such that if you only look at the numbers from there upwards in the matrix, you can form a Latin square out of them.
The minimum is at least 15, because the following selection exists: $$\begin{bmatrix} 11&17&*25*&19 & 16\\ 24& 10 & 13 &*15*&3 \\ 12& 5 & 14 & 2 &*18*\\ *23*& 4 & 1 & 8 & 22\\ 6&*20*& 7 & 21 & 9 \end{bmatrix}$$ So now we blot out numbers $15$ or less. If a group with a larger minimum exists, it must be selected entirely from the below cells: $$\begin{bmatrix} &17&25&19 & 16\\ 24&&&&\\ &&&& 18\\ 23 &&&& 22\\ & 20 && 21 & \end{bmatrix}$$ We see that $25$ must be in the group, as it is the only one in its column. We can remove its row and column to get $$\begin{bmatrix} 24&&&\\ &&& 18\\ 23 &&& 22\\ & 20 & 21 & \end{bmatrix}$$ Now we see that $24$ is the only one in its row. We can do the same procedure: $$\begin{bmatrix} && 18\\ && 22\\ 20 & 21 & \end{bmatrix}$$ The same for $18$ and $20$: $$\begin{bmatrix} \qquad \end{bmatrix}$$ But now we cannot put any more elements in the group, and all our choices were forced. We conclude that the largest minimum is $15$.