Challenge #04  Training at the Academy
Your Answers 
(22/09) Ludercrs : now in the first round it is said that 1 person can fight any number of others. but in the second round he cannot fight the same people. and now if they are devided in groups of 4 then the number of students that could be left over is either 0 or 1. here are the possibilities: student 1 fights student 2, student 3 fights student 4. (0 left over) student 1 fights student 2 and 3. student 4 is out. (1 left over) student 1 fight student 2 and 3 and 4. (0 left over) I hope this is what you were looking for.
Shika : There could be more fighters. Could you generalize this arguement in a way that our readers could understand?

(22/09) kian bung : Let the number of students be x.
Since a person cannot spar himself, the total number of opponents each
student must spar is (x1).
Since the table in round B can become identical with round A just by
changing the student labels, it can be concluded that the total number
of opponents faced by all students combined in round A is equal to that
in round B. With that, we can conclude that there is an even number of
opponents.
If the number of opponents (x1) is even, then x is an odd number. The
two possible remainders for dividing an odd number with 4 is 1 or 3.
Thus, there will definitely be 1 or 3 students left out.
Shika : correct approach but ans is wrong, pay more attention to the intricacy of this argument. :)

(21/09) Vensae Kiarae : There is only 1 possible student left. Shikamaru!! Because he's too busy reading all of his emails to participate in round 4. At least...that's what my family decided on when we tried to solve this.
Shika : then I think I'm smarter then your whole family combined, and it's wrong. :P

(21/09) Amy Louise : i was a little confused by some of the wording of the problem but after looking at the problem i made a few assumptions as to what the wording meant. i hope i wasn't too far off ^_^ to start off, i assume that 'one particular battle' indicated that the problem focuses on one ninja's table. also, i assumed that iruka relabled the students by dividing their names into groups of four on the round A table. my last assumption is that the students left over are the students that were involved in round A, but not round B. now that that's out of the way, i broke the first round down into the formula A + B vs. C. A is one student. the one whose table is in question. B represents the allies they have chosen for the first round. C is the group of opponents they have chosen. in the second round, A moves on to fight B since they haven't faced each other. C, having faced A and B both, is out of this one. the formula for the new round is A vs. B. now we take the numbers into consideration. A vs. B becomes 1 vs. X. since A is one student and B is an unknown number. we know that the formula for the first round is relative since the tables look the same. we'll multiply by 4 since iruka arranged the students into groups of four. the formula becomes 4 vs. 4X. we now know that A + B vs C. is equal to 4 vs. 4X. from that we can determine that A + B = 4. we know that A = 1 so B must equal 3. because in the second round B is X, we now see that X is 3. going back to the first round formula we use that 3 to solve for C. C = 4X. or C = 4*3. further simplified, C = 12. C is the group not participating in the second round and so my answer is 12.
Shika : But 12 is divisible by 4, so I guess you mean zero which is wrong.

(20/09) Alan Nguyen : if group #1 fights every single group there is, then they would not fight anyone in the second combat round, which means there would be 4 people left out for round B
Shika : your answer is half right but I'm not sure of your resaoning

(15/09) Y God : dear shika, Let's say there are N persons.
1 person can fight with all others(N1) and so on...
a team of 2 can fight with all others(N2) and so
on...
a team A(1 person) fight with N1 > left out
a team A(1 person) fight with N2 > left out
a team A(1 person) fight with N3 > left out
a team A(1 person) fight with N4 > left out
a team A(2 persons) fight with N2 > left out
a team A(2 persons) fight with N3 > left out
a team A(2 persons) fight with N4 > left out
a team A(3 persons) fight with N3 > left out
a team A(3 persons) fight with N4 > left out
a team A(4 persons) fight with N4 > left out
possible numbers of students that will be left over:10
Shika : You do know that 10 is bigger that 4 ? Anyway its going to be wrong :P

(15/09) Hatake Kakashi : The answer is 4
Shika : Any number divided by 4 is 0,1,2,3. If this is the answer, I could have asked "what is the remainder of the number that I am thinking of when divided by 4", and do away with that long story right :)

(10/09) Hannah Slade : We think the answer is 3  The 3 anime girlz
Shika : We think the answer is wrong  The 3 amigoz

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