Asif , Aniqa and Saki are perplexed with grid and numbers

Question

Asif has filled in a 3×3 grid with the numbers 1,2, . . . ,9. Saki writes down the three numbers obtained by multiplying the numbers in each horizontal row. Aniqa writes down the three numbers obtained by multiplying the numbers in each vertical column. Can Asif fill in the grid in such a way that Saki and Aniqa obtain the same lists of three numbers? What if the Asif writes the numbers 1,2, . . . ,25 in a 5×5 grid? Or 1,2, . . . ,121 in a 11×11 grid? Can you find any conditions that guarantee that it is possible or any conditions that guarantee that it is impossible for Asif to write the numbers 1,2, . . . , $n^2$ in a n×n grid so that Saki and Aniqa obtain the same lists of numbers?

My observation:

Asif's grid →  1 2 3       Sakis' three numbers= 6 120 504
               4 5 6      Aniqas' three numbers= 28 80 162
               7 8 9

I think it is possible for a magic square if it was addressed as sum. A square array of numbers, usually positive integers, is called a magic square if the sums of the numbers in each row, each column, and both main diagonals are the same.

Then the first grid will be:

Asif's modified grid →  2 7 6  and so Sakis' three numbers= 15 15 15
                        9 5 1        Aniqas' three numbers= 15 15 15
                        4 3 8

I knew a term called multiplication magic square where all number are not needed to be in consecutive. So I can't next what to do?

Or it was asking for it: \begin{array}{|c|c|c|} \hline A & B & C \\\hline D & E & F \\\hline G & H & I\\\hline \end{array}

then,

\begin{align} ABC&=ADG \tag{1}\\ DEF&=BEH \tag{2}\\ GHI&=CFI \tag{3}\\ \end{align}

Please help me the solution.

Thank you so much for your help.

Answer 1

One thing you know immediately is that $7$ and $5$ must be in one of the $A,E,I$ position.

Using the above fact and some quick trials, the following is found to be a valid construction.

\begin{array}{|c|c|c|} \hline 7 & 1 & 6 \\\hline 2 & 5 & 4 \\\hline 3 & 8 &9\\\hline \end{array}

Edit: sorry I overlooked the other two cases asked in the question.

For the $11\times11$ case, there are $13$ primes numbers from $61$ to $121$, namely $61,67,71,73,79,83,89,97,101,103,107,109,113$ must be in the $11$ diagonals so clearly impossible. $5\times5$ seems possible, I will add to the answer if managed to find one.

Update: here's one: \begin{array}{|c|c|c|} \hline 13 & 11 & 24 & 9 & 14 \\\hline 22 &17 &1 & 20 &8 \\\hline 6 &5 & 19& 2 & 18\\\hline 12& 4 & 3 & 23 &25\\\hline 21& 16& 15& 10& 7\\\hline \end{array}

In general, if the number of prime numbers $p$ of the following property:

$p > n$ and $p\leq n^2$ and the number of multiples of $p$ between $n+1$ nad $n^2$ is odd

is greather than $n$ then we can immediately say the $n\times n$ configuration is impossible. This number grows quickly comparing to $n$ (at least at the speed of prime numbers that appears between $n^2\over 2$ and $n^2$ which is $O({n^2\over \ln(n)})$ by Prime Number Theorem which is much larger than $n$). Even when $n=8$ we have $37,41,43,47,53,59,61$ whose multiples appear only once(odd) and $17,19$ whose multiples appear $3$ times(odd) therefore impossible. Smaller cases need to be checked manually.