Is it supposed to be $i-x,j-y\equiv0,-1,1\pmod{30}$ instead?

ABCDE wrote: Is it supposed to be $i-x,j-y\equiv0,-1,1\pmod{30}$ instead? Thank you, yes

(a) Suppose that there are $n$ non-colored squares and $900-n$ colored squares. Consider all ordered pairs of neighboring squares consisting of one colored square and one non-colored square. Every non-colored square can be a part of at most 8 ordered pairs, so there are at most $8n$ of them. Every colored square has to be part of at least 2 ordered pairs as well, so there are at least $2(900-n)$ of them. Hence, $8n\geq2(900-n)\implies n\geq 180$, so there are at most $720$ colored squares. This can be achieved by coloring a square $(i,j)$ if $5\nmid2i-j$. (b) Note that all connected components of colored squares consist of a single square or a domino. Hence, we want to find the maximum number of those we can put down on a 30x30 torus without any of them touching by edge or corner. Consider expanding all of those colored unit squares and dominoes by 1 on the bottom and the left, so they become 2x2 and 2x3 rectangles on the torus. Our original condition of no two colored squares or dominoes becomes no two colored expanded rectangles overlapping. We just want to find the maximum possible value of $2A+B$, where we place $A$ 2x3 rectangles and $B$ 2x2 squares onto a 30x30 torus without any rectangle overlapping another. We clearly have $6A+4B\leq 900$, so $2A+B=\frac{6A+4B}{3}-\frac{B}{3}\leq\frac{6A+4B}{3}=300$. This is achievable by coloring all squares with $2\nmid i,3\nmid j$. Can someone check the above? This feels too simple for an Iran final round #3

such that any colored square has at most $k$ neighbors is it mean??

ABCDE wrote: We clearly have $6A+4B\leq 900$ I think this part is wrong. Can have squares from outside of the table Or am I wrong? (I've an example with 310 instead of 300)