Take a $n \times n$ chess page.Determine the $n$ such that we can put the numbers $1,2,3, \ldots ,n$ in the squares of the page such that we know the following two conditions are true: a) for each row we know all the numbers $1,2,3, \ldots ,n$ have appeared on it and the numbers that are in the black squares of that row have the same sum as the sum of the numbers in the white squares of that row. b) for each column we know all the numbers $1,2,3, \ldots ,n$ have appeared on it and the numbers that are in the black squares in that column have the same sum as the sum of the numbers in the white squares of that column.
Problem
Source: Iran 2nd round 2022 P3
Tags: combinatorics, table
th1nq3r
09.05.2022 19:29
Pretty sure $n$ is a multiple of $4$...
Pmshw
09.05.2022 19:30
th1nq3r wrote: Pretty sure $n$ is a multiple of $4$... It Should be… I’ll probably post my solution tomorrow…
MathSaiyan
09.05.2022 23:45
As @above said, answer is only multiples of 4. Construction left to reader .
Main idea: color the board modulo 2. If lower left is $(0,0)$, then paint $(0,0)$, $(1,0)$, $(0,1)$ and $(1,1)$ with colors $M$, $A$, $B$ and $m$, respectively (paint the remaining cells cyclically mod 2). For some $i\in \{1,2,\ldots,n\}$, denote $i_{\text{color}}$ to be the number cells of that color with number $i$.
Observe that the sum of the cells of each color is the same for each of the four colors, thus $4\mid n\cdot\frac{n(n+1)}{2}$. Thus we discard the $n\equiv 2$ (mod 4) case.
Now assume $n$ is odd.
Lemma:
$i_M = i_m+1$ for all $i=1,2,\ldots,n$.
Proof:
Algorithmical. Suppose we start with all $i$'s placed at a main diagonal, then see what happens when we invert the corners of some rectangle with two opposite corners with number $i$. It's easy to see the lemma still holds after the change.
Using this, the fact that the sum of numbers in $M$-cells is the same as the sum of the numbers in $m$-cells easily gets to contradiction.
alinazarboland
10.05.2022 06:57
The answer is the numbers divisible by 4. The example can be found easily. Clearly $1+2+...+n$ is even so $n(n+1)$ is divisible by $4$. So either $n=3$ or $n=0$ $mod 4$. Assume FTSOC $n=3$ mod 4 for some $n$ , satisfies in the condition. Color the columns alternatively red and blue with the left-most and right-most column red. Clearly the sum of numbers in the red cells is more than the sum of numbers in blue ones since tbe sum of all numbers in a column is a constant value and there are more red columns than blue columns. On the other hand , in every row : ${blacks}={reds} and {whites}={blues}$ or ${blacks}={blues} and {whites}={reds}$. So the sum of red cells is equal to sum of blue cells in every row. So the sum of red cells is equal to the sum of blue cells in the table. Contradiction. So the only answer is $n=0$ mod 4
Titusir
03.10.2022 04:29
MathSaiyan wrote:
As @above said, answer is only multiples of 4. Construction left to reader .
Main idea: color the board modulo 2. If lower left is $(0,0)$, then paint $(0,0)$, $(1,0)$, $(0,1)$ and $(1,1)$ with colors $M$, $A$, $B$ and $m$, respectively (paint the remaining cells cyclically mod 2). For some $i\in \{1,2,\ldots,n\}$, denote $i_{\text{color}}$ to be the number cells of that color with number $i$.
Observe that the sum of the cells of each color is the same for each of the four colors, thus $4\mid n\cdot\frac{n(n+1)}{2}$. Thus we discard the $n\equiv 2$ (mod 4) case.
Now assume $n$ is odd.
Lemma:
$i_M = i_m+1$ for all $i=1,2,\ldots,n$.
Proof:
Algorithmical. Suppose we start with all $i$'s placed at a main diagonal, then see what happens when we invert the corners of some rectangle with two opposite corners with number $i$. It's easy to see the lemma still holds after the change.
Using this, the fact that the sum of numbers in $M$-cells is the same as the sum of the numbers in $m$-cells easily gets to contradiction.
Can u explain your colouring idea once again
skyboy2007
25.12.2022 17:45
n(N+1)/2=2K SO N DIVIDES 4 or n+1 divides 4 you can easily see it is wrond for odds so the only answer is n which is divides 4