Hello . tell me if there is mistake in mu solution _____ first claim : Alice can colour 100 squares and Bob can colour the entire board . name the rows$ A_1,A_2,A_3,...A_{100}$ and name the columns $B_1,B_2,B_3...,B_{100}$ then if Alice colour $ A_1B_1,A_1B_2,...,A_1B_{10}, A_2B_1,A_2B_2,..A_2B_{10} ,...A_{10}B_{10} $Bob can colour the entire board ____ second claim : Alice has to colour at least 100 squares proof : suppose the number of squares is less than 100 WLOG , suppose Bob starts with colouring the rows . at first , He colours $A_1$ rows which means there are at least $10A_1 $red squares . then , If $A_1<10$ , He colours$ B_1$ columns which means there are at least $B_1(10-A_1)$ more squares . if $B_1<10$ , He colours $A_2$ and there are at least $A_2(10-B_1)$ more and if $A_1+A_2<10$ He colours $B_2$ and there are at least $B_2(10-A_1-A_2 )$ ... at the $k_{th}$ step if either $A_1+..+A_k=10$ or $B_1+...B_t=10$ and He finishes colouring . now , It is easy to show that He can colour the rest of the squares . As we have assumed , $10A_1+B_1(10-A_1)+A_2(10-B_1)+...Bt(10-A_1-A_2-...-A_k)$ <= the number of the squares < 100 so , $10(A_1+...+A_k+B_1+...+B_t) - (A_1+...A_k)(B_1+...B_t)$ < 100 and $A_1+...A_k=10 $or $B_1+...B_t=10$ . WLOG assume that$ A_1+...A_k=10 $ so , 100 <100 which is contradiction

Clearly, Bob can do the job when $k=100.$ Alice initially colors a square $10\times 10$ and Bob colors the entire board. We'll prove it cannot be done in case $k<100.$ Consider a more general problem. Let $a,b\in \mathbb{N}.$ Bob can cover an entire row if it has at least $a$ cells on it or a column if it contains at least $b$ cells. We are looking for a minimum $k=k(a,b)$ such that it's possible to color the entire board. As expected $k=ab.$ Note that the size of the board doesn't matter, as long as it's larger than $a$ and $b.$ We can assume we have an infinite board. It means that if we can color a finite board we can color also an infinite board and vise versa. Consider the first move of Bob. He either colors a row or a column. Assume he colrs a row. In this case, we have at least $a$ cells on it that were colored initially. In each subsequent move Bob's environment is as if he were playing the same game but with the pair $(a,b-1)$ instead of $(a,b).$ Indeed, at each following step the legal moves of both games are the same. It means $$k(a,b)\ge a+k(a,b-1).$$In case Bob's first move is to color a column, the same argument yields $$k(a,b)\ge b+k(a-1,b).$$Obviously, $k(1,1)=1$ which gives us $k(a,b)\ge ab.$ Hence, $k(a,b)=ab.$

I would like to solve this problem generalized for $k^2 \times k^2$ and $10\to k$ Notice that we shall color at least $k$ squares that they all lie on same row or column(it doesnt matter as we can obtain the second one by flipping the board).Then second player colors that row when the first player coloured already $R_1$ then we can see that second player can not make a move and he needs $k-1 \times k$ squares colored to obtain a full colored board.When coloring $2$ rows we need $k-2$,when we color $3$ we need $k-3$ etc…$k-a$ for $a$…but when we finally color $k$ rows we need $0$ so it is a valid number.Now we prove it is the smallest.If there is a number $k-z$ such that it is valid.If there is $mk$ $k$ colored rows we can full-color $m$ rows,after that to color a new col-row we need to use the the ones we colored previously,but we need $k-m$ but we do not have them so $k$ is the answer