Finding the largest minimum 
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?
 A: 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 $15$.
