Solutions

1) (un)Lucky Luke warm up: L(99; 6). Consider the set of 7 tickets [1 2 3 4 5 6], [7 8 9 10 11 12], [13 14 15 16 17 18], [19 20 21 22 23 24], [25 26 27 28 29 30], [31 32 33 34 35 36], [37 38 39 40 41 42]. Since the 7 tickets are disjoint, no matter what 6 numbers are drawn, at least one of the tickets will not include any of the drawn numbers. Thus, L(99; 6) is at most 7. We can also see that 6 tickets is not enough. No matter what choice of 6 tickets is chosen, it is possible for one number from each ticket to be drawn when selecting a winner. So L(99; 6) = 7.