Problem 2

Find all integers

n

for which each cell of an

n\times n

table can be filled with one of the letters

I

M

, and

O

in such a way that: 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 on that diagonal are

I

, one third are

M

, and one third are

O

. (Note that an

n\times n

table has

4n-2

diagonals in total, in both directions.)

Step 5 of 5: Conclude

9\mid n

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In plain words

The global count of $I$ agrees with the two-way count only when $n^2/9$ is divisible by $3$ .

\tfrac13n^2=\tfrac49n^2-3a\quad\Rightarrow\quad 9\mid n

Detailed analysis

The row condition gives exactly $n^2/3$ cells containing $I$ . Equating this with the count from Steps 3 and 4 gives $n^2/3=4n^2/9-3a$ , hence $3a=n^2/9$ . Thus $3\mid n^2/9$ , and since $3\mid n$ , this is equivalent to $9\mid n$ . Together with the explicit periodic tile, the answer is exactly the positive multiples of $9$ .