-4
$\begingroup$

I find a interesting question of Prmo mock and Promys 2020

For which $n\in\mathbb{N}$ is it possible to arrange $\{1,…,n^2\}$ in an $n\times n$ grid so that the set of products of columns equals the set of products of rows?

I can find solution for $3\times 3$ or $4\times 4$ but unable to generalize for $n\times n$.

The question has been answered by Rob Pratt but I am unable to understand his solution .

Can anyone please help me in this question

$\endgroup$
3
  • $\begingroup$ @JosephVanName sir , is there proof of this , pls share $\endgroup$ Commented Feb 11, 2023 at 12:41
  • 4
    $\begingroup$ Welcome to MathOverflow! This is a question for professional mathematicians to ask questions about their research, as is explained in the help centre. Maths Olympiad problems are almost certainly off topic here, but you could try asking on Maths Stack Exchange. $\endgroup$ Commented Feb 11, 2023 at 13:17
  • $\begingroup$ @R.vanDobbendeBruyn i didn't knew sorry $\endgroup$ Commented Feb 11, 2023 at 14:20

1 Answer 1

2
$\begingroup$

Here is the solution of the Promys 2020 problem, following RobPratt and Mike Earnest at Can we arrange $\{1,...,16\}$ in $4\times 4$-grid so {products of rows} = {products of columns}?.

Theorem. For $n\in\{1,3,4,5,6,7,8,10\}$ one can arrange $\{1,\dotsc,n^2\}$ in an $n\times n$ grid so that the set of products of columns equals the set of products of rows. For $n\notin\{1,3,4,5,6,7,8,10\}$ there is no such arrangement.

Proof. For $n\in\{1,3,4,5,6,7,8,10\}$, see the constructions of RobPratt. For $n=2$ it is easy to check by hand there there is no arrangement. Now assume that $n=9$ or $n\geq 11$. Then, following Mike Earnest's great idea, it suffices to show that $\pi(n^2)-\pi(n^2/2)>n$. For $n\in\{9,11,12,13,14,15,16\}$ we can verify this by hand. For $n\geq 17$ the required inequality follows from Corollary 3 in Rosser–Schoenfeld: Approximate formulas for some functions of prime numbers. Indeed, by this theorem we have that $$\pi(n^2)-\pi(n^2/2)>\frac{(3/10)n^2}{\log(n^2/2)}>n,\qquad n\geq 17.$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .