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 4 of 5: Count the remaining

I

-cells by rows and columns

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In plain words

Rows and columns indexed $2$ modulo $3$ contribute the complementary count; their overlap is again the good $I$ -cells.

|C\cup R|=\tfrac29n^2-a,\quad |I|=\tfrac49n^2-3a

Detailed analysis

Let $C$ be the $I$ -cells in columns $3s+2$ , and $R$ the $I$ -cells in rows $3r+2$ . Each selected column and row is balanced, so $|C|=|R|=n^2/9$ ; their intersection is the $a$ good $I$ -cells. Thus $|C\cup R|=2n^2/9-a$ . The cells counted in exactly one diagonal family contribute $2n^2/9-2a$ , while the row/column union contributes $2n^2/9-a$ , so the total number of $I$ -cells is $4n^2/9-3a$ .