- in each row and each column, one third of the entries are $I$, one third are $M$ and one third are $O$; and
- in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are $I$, one third are $M$ and one third are $O$.
IMO 2016 Problem 2
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
Find all integers $n$ for which each cell of $n \times n$ table can be filled with one of the letters $I,M$ and $O$ in such a way that:
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- ahmedittihad
- Posts:181
- Joined:Mon Mar 28, 2016 6:21 pm
Re: IMO 2016 Problem 2
\[
\begin{array}{|ccc|ccc|ccc|}
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
\end{array}
\]
I could only show that $9$ divides $n$. No idea about the other part. How many points would that give me?
\begin{array}{|ccc|ccc|ccc|}
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
\end{array}
\]
I could only show that $9$ divides $n$. No idea about the other part. How many points would that give me?
Frankly, my dear, I don't give a damn.
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: IMO 2016 Problem 2
Here is a proof that $9|n$ is necessary (see ahmedittihad's post for sufficiency).
It's obvious that $n=3k$. Now, Replace $I$ with $1$, $M$ with $\omega$ and $O$ with $\omega^2$. Now, sum the numbers in the:
1. Rows that are of the form $3m+2$.
2. Columns that are the form $3m+2$.
3. All diagonals of both types that have number of cells divisible by $3$.
Note that while all of them are $0$, The sum of whole grid is $0$ and summing the three of them would result in the sum of the whole board plus twice the sum of the central squares when the whole grid is divided into $3\times 3$ parts. So, their sum is also 0. Assume that there are $a$ $I$'s, $b$ $M$'s and $c$ $O$'s in the central squares. So, we get:
$$a+b+c=k^2$$
$$a+b\omega+c\omega^2=0$$
Implying that $(a-b)^2+(b-c)^2+(c-a)^2=0$, From which we get $3|k$ and so $9|n$.
It's obvious that $n=3k$. Now, Replace $I$ with $1$, $M$ with $\omega$ and $O$ with $\omega^2$. Now, sum the numbers in the:
1. Rows that are of the form $3m+2$.
2. Columns that are the form $3m+2$.
3. All diagonals of both types that have number of cells divisible by $3$.
Note that while all of them are $0$, The sum of whole grid is $0$ and summing the three of them would result in the sum of the whole board plus twice the sum of the central squares when the whole grid is divided into $3\times 3$ parts. So, their sum is also 0. Assume that there are $a$ $I$'s, $b$ $M$'s and $c$ $O$'s in the central squares. So, we get:
$$a+b+c=k^2$$
$$a+b\omega+c\omega^2=0$$
Implying that $(a-b)^2+(b-c)^2+(c-a)^2=0$, From which we get $3|k$ and so $9|n$.
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.