IMO Team Selection Contest II

a) Let's fill the top left $(n-1)\times (n-1)$ subtable arbitrarily; this can be done in $6^{(n-1{)}^2}$ ways. Now there are $3$ ways to fill each of the top $n-1$ cells of the rightmost column and $2$ ways to fill each of the left $n-1$ cells of the bottom row to satisfy the requirements. The value for the last empty cell in the bottom right is then uniquely determined (mod $2$ by the bottom row, and mod $3$ by the rightmost column). In conclusion, there are $6^{(n-1{)}^2}\cdot 3^{n-1}\cdot 2^{n-1}=6^{n^2-n}$ ways to fill the table.

b) For $n=1$, the only solution is writing $0$ into the single cell. For $n=2$, let $a$ be the top left number. The bottom right must then be $(6-a)$ mod $6$. Using the conditions for rows and columns, for the top right number $x$ we get the equations $x\equiv -a(\mathrm{mod}2)$ and $x\equiv a(\mathrm{mod}3)$, and for the bottom left number $y$, $y\equiv a(\mathrm{mod}2)$ and $y\equiv -a(\mathrm{mod}3)$. The Chinese remainder theorem determines $x$ and $y$ uniquely, and we see from the equations that their sum is also divisible by $6$. Thus there are $6$ ways to fill the table in this case, one for each value of $a$.

Consider now $n\ge 3$. Fill the top left $(n-1)\times (n-1)$ subtable arbitrarily; this can be done in $6^{(n-1{)}^2}$ ways. The bottom right cell's value is uniquely determined by other values on the falling diagonal. Denote the value in the top left cell by $a$, the sum of the $2$nd to $(n-1)$-st cells (inclusive) in the top row by $b$, the sum of the $2$nd to $(n-1)$-st cells in the leftmost column by $c$, and the sum of $2$nd to $(n-1)$-st cells on the rising diagonal by $d$.

Using the Chinese remainder theorem, fill the top right cell with the unique value $x$ such that $x\equiv -a-b(\mathrm{mod}2)$ and $x\equiv a+c-d(\mathrm{mod}3)$, and the bottom left cell with the unique value $y$ such that $y\equiv a+b-d(\mathrm{mod}2)$ and $y\equiv -a-c(\mathrm{mod}3)$. The divisibility conditions are now fulfilled for the top row, the leftmost column and both diagonals (the rising diagonal is verified by summing mod $2$ and mod $3$ separately).

Now, we leave one cell both in the rightmost column and in the bottom row empty for the time being. For the other $n-3$ empty cells in the rightmost column, there are $3$ possible values for each, and for the other $n-3$ empty cells in the bottom row, $2$ values for each. Having made all those choices (which can be done in $3^{n-3}\cdot 2^{n-3}$ ways), the values for the two remaining cells are now uniquely determined (mod $2$ by the values in the respective row, and mod $3$ by the column). The total number of ways to fill the table is $6^{(n-1{)}^2}\cdot 3^{n-3}\cdot 2^{n-3}=6^{n^2-n-2}$.