Post Number:**#3** by **Thanic Nur Samin** » Mon Jan 02, 2017 1:54 pm

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 $9|n$.

Hammer with tact.

Because destroying everything mindlessly isn't cool enough.