IMO D
A = tickets sold by the first member
B = tickets sold by the second member
C = tickets sold by the third member
A >=0, B >=0, C >=0
1) A + B + C = 6
if you don't want any of the number to be at least 2, all the numbers have to be <= 1
is it possible that A + B + C = 6 and 1>= A >= 0, 1>= B >= 0, 1>= C >= 0? No so statement 1 is sufficient
2) No 2 members sold the same number of tickets
let's start assigning values and try to prove that all can be <= 1
A = 0
for B, we cannot use neither 0 nor >=2 so B has to be 1 B =1
we run out of possibilities for C, so C has to be at least 2 sufficient
