Did one of the three members of a certain team sell at least 2 raffle tickets yesterday?

(1) the 3 members sold a total of 6 raffle tickets yesterday
(2) No 2 of the members sold the same number of raffle tickets yesterday

During the test, I was confused with the way the problem was written. I did not clearly understand what "did one of the three" meant.

Anyways, I am thinking the question is asking if at least one of the three sell at least 2 tickets. With this assumption, I think this is the solution. Tutors, please correct if wrong

(1) if they sold 6 together, the possibilities (2,2,2), (1,2,3), (0,3,3) (different variations of these). In all cases, there is at least one with 2 or more.

(2). This I think is real cool.. if one of them is 0, the other is 1, the third one has to be 2 or more, hence sufficient.

Hence the answer is D.

I got this wrong in the exam.

-Raj.

1) Sufficient as even if 1 member sold 0 ticket, one member must have sold atleast 2 tickets as total sold tickets is 6.
2) Not sufficient because number of tickets sold is not given.

So answer should be B.
What is the correct answer.

Answer is D.

for 2), this is sort of tricky.. even if two of them sold the lowest possible 0 and 1, the third one has to sell at least 3 since they all sold different number of tickets. You dont actually have to know the total number. Hope that clarifies..
-Raj.

yeah, this should be (d).

statement (1):
there's a statement called the pigeonhole principle, which basically says the following two things:
* if the AVERAGE of a set of integers is an INTEGER n, then at least one element of the set is > n.
* if the AVERAGE of a set of integers is a NON-INTEGER n, then at least one element of the set is > the next integer above n.
this principle is easy to prove: if you assume the contrary, then you get the absurd situation in which every element of a set is below the average of the set. that is of course impossible.

specifically, statement (1) is a case of the first part of the principle: the average of the set is 6/3 = 2, so at least one element of the set must be 2 or more.
again, you can prove this by reductio ad absurdum: if no one had sold 2 or more tickets, then you'd have a set in which everyone sold either 0 or 1 ticket, but the average is somehow still 2. that's untenable.

--

statement (2):
there are only two ways not to sell at least 2 tickets: sell 0 tickets, and sell 1 ticket.
if everyone sells a different # of tickets, then you can't fit three people into these two categories.
therefore, someone must have sold at least 2 tickets.