Challenge #04  Training at the Academy
Your Answers 
(25/10) Gekido Tokumei : the answer is 0
Shika : please refer to previous posting.

(14/10) Rekka Hanabishi : Shika, the number of nin's left over is 1. First of all, for the tables to be similar, the number of opponents that each individual fights is the same for round A and B. If we let the number of nins be n, the number of nins that each nin has to fight in each round is (n1)/2. Since they have to be equal, n1 must be an even number. Hence, the only possibilities are 1 and 3. Let's say A chooses to fight B, B will have no choice but to fight A. As a result, when you tabulate the tables, the number of cells will always be even. Considering the table, the number of rows(number of nins) it has is n, and the number of columns(opponents) is (n1)/2. Thus, the total number of cells is n(n1)/2. Since it has to be even, (n1)/2 has to be even(n1 is even, n is odd, odd x odd = odd. even x odd = even). Since (n1)/2 is even, n cannot be 7, 11, 15... hence, the remainder cannot be 3. So the number of nin's left over is 1.
Shika : Your first assertion is questionable, that is why you didn't get it

(13/10) Spray M727 : I think that the answer is 0.
Shika : I would prefer some form of reasoning, but any way your answer is wrong

(10/10) E.S.M. : I've taken the liberty of pasting the original question so that this would be easier to comprehend. In your question you have seemed to add lots of unnecessary information. Your real question never states the original amount of students. The only answerable question was: If the unknown amount of students were divided into groups of four then what would be the possible number of students that would be left over? If you divide a number by four the only remainders possible would be zero, one, two, or three. If you had more than four you would be able to make another set, if you had four you would have another set, and since the amount of students was not given the only possible answers are 0, 1, 2, or 3.
 From the Irish Samurai, and the evil shadow monster who hides in the corner of people's rooms (or you could just call me E.S.M. for short)
Shika : Why would I ruin my own reputation by asking such an idiotic question.

(09/10) Tim Watson : If your referring to Iruka's actual class than there must be 21 students. Divide that by 4 and you get one student left out. My guess is 1. I have an equation, but it's too troublesome to type out.
Shika : Maybe this is sometime in the future and maybe more people has enrolled or maybe the students has kids who become students... anyway wrong answer. Next!

(06/10) Tim Watson : If your referring to Iruka's actual class than there must be 21 students.
Divide that by 4 and you get one student left out. My guess is 1. I have an
equation, but it's too troublesome to type out.
Shika : Maybe this is sometime in the future and maybe more people has enrolled or maybe the students has kids who become students... anyway wrong answer. Next!

(04/10) Atsushi : You can't figure out the answer. You aren't given the number of students in the first place. Unless you are assuming we know how many students are in Iruka's class, we can't answer the problem.
Shika : Refer to the previous try.

(04/10) Jeff : Shikamarusama, I noticed one key component missing from your problem. You never give the size of Iruka Sensai's class, so we can't determine how many students would be left over, because we don't know how many students there are to begin with. I will, however, continue to work on the problem, and I can't wain tuntil you post the answer, and a new question. Thanks, and YOU TOTALLY ROCK! Shikamaru forever!!!
Shika : If I told you the number of students, then this would be a problem on long division which is beneath Shikamaru :P Thanks for your overwhelming support though :)

(03/10) Nara Shikamaru : Hi, this is my answer.
Suppose the number of students in the class is x. If Iruka can just relabel the students for second round, means that each student fights with the same number of students in both the 1st and 2nd round.
Therefore a student fights with (x1)/2 number of students in each round.
Since (x1) is divisible by 2, (x1) is an even number.
So therefore x is an odd number.
If x is an odd number and is divided by 4, then the possible number of
students left over will be either 1 or 3.
Shika : you are wrong but your apporach is similar to the other guy.

(02/10) Konoha Lau : answer can be 1,2,3 its depending on your explanation my explanation is that and y 0 is not included because if its 0 the total will be even,and if total is even,both the tables won't be identical
if total = 41
1(a student) fights 20 (other students) next round he can fight 20 more students remainder if divided by 4 is 1
if remainder is 2 same concepts and total will be 42 1 fights 21
nxt round another 21(identical as round A)
and 3 is the same...so...1,2,3 can be correct right?
Shika : any number divided by 4 would give you 1,2, or3

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