seems like it is 7, mark 7 out of 10 cells on the first row example: 1 0 1 1 0 1 1 0 1 1 . Where 1 is marked cell and 0 is not marked.

amirthenumb14 wrote: seems like it is 7, mark 7 out of 10 cells on the first row example: 1 0 1 1 0 1 1 0 1 1 . Where 1 is marked cell and 0 is not marked. Do you have a proof that there is no example with less than 7 cells marked?

so we must mark at least 1 cell, soo lets mark only a row of 3*10 cells and others we leave without marked cells also we will mark only cells on top of the table, obviously we will minimize out amount of marked cells, so we have 3 3*3 squares so we will mark 6 cells and also we have a 1*3 row so we need to mark another 1 cell, and get our answer. sorry for my quite bad explanation, my English is way too bad(((

amirthenumb14 wrote: so we must mark at least 1 cell, soo lets mark only a row of 3*10 cells and others we leave without marked cells also we will mark only cells on top of the table, obviously we will minimize out amount of marked cells, so we have 3 3*3 squares so we will mark 6 cells and also we have a 1*3 row so we need to mark another 1 cell, and get our answer. sorry for my quite bad explanation, my English is way too bad((( I don't understand why marking cells only in the top $3 \times 10$ subtable is optimal, and why can't you have $0$ in any of the three $3 \times 3$ subtables or the $1 \times 3$ subtable