A 5×5 board has $25$ cells. The numbers $\{1,2,3,4,5\}$ are written on every row,every column and the two main diagonals without any repetition. If the sum of the numbers of the diagonal below the leading diagonal be represented as $A$. Show that $A\ne 20$. Also find the maximum possible value of $A$.
I could prove that $A\ne 20$ by showing that each of the cells of the diagonal below the leading diagonal cannot be $20$ as if such arrangement exists, then the board cannot have $5$ five times, which actually should be by the given condition of the question.
For the second part, I can imagine the maxiumum value to be $17$. For which I proved that $5$ can occur only $2$ times in the four cells and probably $4$ cannot occur twice. But I got no formal proof. Please help!
Best Answer
There can be at most $2$ equal numbers on subdiagonal:
$$ \begin{array}{|c|c|c|c|c|} \hline .&.&.&.&. \\ \hline A&.&.&.&.\\ \hline .&A&.&.&.\\ \hline .&.&.&.&.\\ \hline .&.&.&.&. \\ \hline \end{array} \implies \begin{array}{|c|c|c|c|c|} \hline .&.&.&.&A \\ \hline A&.&.&.&.\\ \hline .&A&.&.&.\\ \hline .&.&.&A&.\\ \hline .&.&A&.&. \\ \hline \end{array}; $$
$$ \begin{array}{|c|c|c|c|c|} \hline .&.&.&.&. \\ \hline A&.&.&.&.\\ \hline .&.&.&.&.\\ \hline .&.&A&.&.\\ \hline .&.&.&.&. \\ \hline \end{array} \implies \varnothing; $$
$$ \begin{array}{|c|c|c|c|c|} \hline .&.&.&.&. \\ \hline A&.&.&.&.\\ \hline .&.&.&.&.\\ \hline .&.&.&.&.\\ \hline .&.&.&A&. \\ \hline \end{array} \implies \begin{array}{|c|c|c|c|c|} \hline .&A&.&.&. \\ \hline A&.&.&.&.\\ \hline .&.&A&.&.\\ \hline .&.&.&.&A\\ \hline .&.&.&A&. \\ \hline \end{array}; $$
$$ \begin{array}{|c|c|c|c|c|} \hline .&.&.&.&. \\ \hline .&.&.&.&.\\ \hline .&A&.&.&.\\ \hline .&.&A&.&.\\ \hline .&.&.&.&. \\ \hline \end{array} \implies \varnothing; $$
(other cases are symmetrical).
Other $2$ cells cannot be filled with $2$ equal numbers, because we will get collision in red cell:
$$ \begin{array}{|c|c|c|c|c|} \hline .&.&.&.&. \\ \hline A&.&.&.&.\\ \hline .&A&.&.&.\\ \hline .&.&B&.&.\\ \hline .&.&.&B&. \\ \hline \end{array} \implies \begin{array}{|c|c|c|c|c|} \hline .&.&.&.&\color{red}{A,B} \\ \hline A&B&.&.&.\\ \hline B&A&.&.&.\\ \hline .&.&B&A&.\\ \hline .&.&A&B&. \\ \hline \end{array}. $$
So, subdiagonal has either $4$ different numbers, or $2$ equal and $2$ different.
$4$ different numbers can reach only value of $14$: $5+4+3+2$.
$2$ equal and $2$ different numbers can reach value of $17$: $5+5+4+3$, and no more.
Examples:
$4\;3\;2\;1\;5 ~~~~~~~ 1\;5\;4\;3\;2$
$5\;2\;1\;4\;3 ~~~~~~~ 5\;3\;2\;4\;1$
$1\;5\;3\;2\;4 ~~~~~~~ 2\;4\;5\;1\;3$
$3\;1\;4\;5\;2 ~~~~~~~ 4\;1\;3\;2\;5$
$2\;4\;5\;3\;1 ~~~~~~~ 3\;2\;1\;5\;4$