Your Answers ----------------------------------------
(20/08) Ryuu Hime : First, to state the obvious: there will be either 0, 1, 2, 3 students left over. There can't be more than that, because then you would be able to form another squad.
Say that there are X students in this particular class. Each student would then have to fight X-1 students total. (Assume that you cannot fight yourself. ^.~) In order for the groups to be the same after the re-labelling, the number of other students fought by a particular student must be equal for both rounds. Thus, the number of students fought in a round can be represented by the expression (X-1)/2. Since (X-1)/2 must be a whole number (no such thing as half a ninja), it follows that X-1 is an even number (any whole number multiplied by 2 is even). Therefore, X, the number of students, must be odd. When dividing students into squads of four, there will be either one or three students left over.
P.S. For some reason this seemed too easy...
Shika : One can benefit much from doubt. Don't be so sure of yourself. Next.
(20/08) Lady Nyoko : First of all, for Round A, we must assume that once a student has chosen an opponent, the opponent also has the chance to, in turn, choose their own opponent. With this reasoning, it is possible to have groups of over 3 students fighting together. For example...
Student 1 chooses to fight Student 4
But Student 4 chooses to fight Student 3
And Student 3 chooses to fight Student 2
Finally, Student 2 chooses to fight Student 1
Therefore, with this reasoning, Students 1 and 3 are grouped together against Students 2 and 4.
With this reasoning, the students have each chosen their own opponent, and are in groups of four.
Now, for Round B, we must assume that the students must spar with new opponents. Since they can't spar with the same student as in their group during Round A, the chart would have to look like this. I am once again, using my example of 4 different students.
Student 1 vs. Student 3
Student 4 vs. Student 2
This way, each student has a new match with an opponent they haven't chosen to fight yet.
Because in Round B, the Students are matched up against only one other opponent, the only possible remaining student would be 1. The reason being, even if there is an odd number of students, if 3 are left over, 2 of those 3 could still be matched up against each other. Therefore, the only correct answer is 1.
Shika : First of all your answer is wrong. Secondly your answer doesn't seem to be exhaustive. The purpose of reasoning is to assault a problem by breaking it down into bite-size logic, that one can have no doubt.
(14/08) CJ : I hope this is the right answer.
4, because in Round A:
1 can fight all but 3 people (people in their squad 1 + 3 = 4). And no matter what amount of people one person fights, their will always be 4 people that can not fight each, so they will be the only remaining people.
This doesn't necessarily matter, but for answer's sake, it doesn't really matter what number is assigned. Naruto can be assigned number 1 and Orochimaru can be assigned number 5, if you just swithc their numbers you get teh same opponents. So that's why I think Round B is extraneous.
There you have it. It sounds better in my head, but I tried my best. If that's not it, do we get a hint? lol.
Shika : Wrong. As for a hint, isn't it more fun this way. But rest assured that answer is within your grasp if you contemplate :)