Question 619582
I'll assume two contestants, A and B.


First off, if contestant A gets 15, 20, 24, or 28, he automatically wins and B's number is 1.


If A gets 6, 7, 8, 10, 12, or 14 (all of these numbers are factors of exactly one of the four given #s), A wins, and B's number will be either 2, 3, or 4.


If A gets any other factor of the given four #'s, he cannot immediately determine B's number.


Hence, if A gets 2, B can have either a 10, 12, or 14. B wins.


If A gets 3, B can have either 5 or 8. If B does not automatically respond with A's number, B cannot have 8, and therefore contestant A can conclude that B has 5.


If A gets 4, B can have either 5, 6, or 7. If B does not automatically respond with A's number, he cannot have 6 or 7. Therefore A can conclude that B has 5.


If A gets 5, B can have either 3 or 4. A cannot determine what number B has, so B wins by the above logic.


Therefore the possible losing numbers are 1, 2, and 5. Other than 1,2,3,4,5, no other numbers divide two or more of {15, 20, 24, 28}.