"Math lotto" is played as follows: a player marks six squares on a 6x6 square. Then six "losing squares" are drawn. A player wins if none of the losing squares are marked on his lottery ticket.
1)Prove that one can complete nine lottery tickets in such a way that at least one of them wins.
2)Prove that this is not possible with only eight tickets.
My attempt is as follows;
Now we consider the case where each rectangle has one cross each.
Now take the two rectangles on the top left of the square (figure 2). These have a total of two crosses.
The first two columns together contains one cross and the third and fourth columns together contains one cross.
There are four cases and we need at least four tickets (ticket 7 to ticket 10) to ensure win.
I am only getting a minimum of ten tickets. How do I prove only nine tickets is required and for eight tickets it is not possible?
Reference: Combinatorics by Stephan Wagner, Page 42, Problem 49 . https://math.sun.ac.za/swagner/Combinatorics.pdf