
Challenge #04  Training at the Academy
Your Answers 
(12/12) Eric Ferraro : The answer to the IQ challenge #5 can range in any multiple of 4. Logic tells you that if he devides the remainder into groups of 4s then there has to be an answer with a multiple of 4 students.
Shika : This defeats the purpose of the story I cooked up. Next!

(10/12) Klave : well, first off...you are right...my first assertion was wrong...it couldn't be even just because the list was identical...and I found another way to look at it...Firstly...they were already divided into
4 squads...which is something i don't keep my mind to so I'm writing it...and a student or a few is left behind so ranging from 13 students...and it didn't have to be identifical if I was to raise the arguement that you said "could be identical if the students were relabeled"...and where am i going in this...back to the drawing board...
back to finding out the number of students...the total number of students was not only even it could also be odd...
the list could be made identical if and only if all the students gang up on one...meaning all groups of four gang up on one student...and there you have it my answer...baaah!!! my first answer was probably more
interesting...maybe not...i'm guessing this... Oh yeah...if there were two or three being ganged up by many...then the list couldn't be identical anymore as they will have to spar each other.I think I'm still wrong though...Well...here it goes
Shika : You're right. You're still wrong. Sorry, but your answer is not exhaustive and lack of justification. :)

(09/12) Zulhilmi Zainuddin : I think the answer will be 4. reason: The number of the genins should be even, through my understanding. We divide them into 2 teams, first team fight with the second team. They can be any number (team 1 can be 55, team 2 can be 45). Because they can't team up with genins who they have already fought, generally they will be team up with they own group, then if possible with the other groups.
I don't know how to elaborate more but this is my equation: a+b/4 = y.
(a=team1, b=team2, y=total team). I think you can get it from my equation, where the 4 people left are those who have fought with each other, so they can't be teammates. In case my assumption is wrong, that the number is not even (I hope not), then possible
answer will only be 1,2,3.
Shika : Your assumption is wrong.And you are basically suggesting all possible remainders, which defeats the purpose of the question.

(07/12) Lynn Kang : The answer is 1.
Shika : Incorrect. Looks like I'll be forced to reveal the answer soon.

(07/12) KingR@ : The correct answer is very simple......your mama!!! LOL, I am done.
Shika : My mother didn't take part in the training. Next!

(07/12) Djamin schurr : Hard to explain because im dutch but..
the battles A and B are already decided when he devides them in groups of 4 so the only logical options are 0,1,2 and 3
but if there are two rounds and everyone can choose their opponents in the first round but eventually have to fight ALL the others in the second round then no matter how many acadamy students there are.. it must be divided by 2
now if you for example pick a number like 15.. the students will be fighting 7 in the first round and in the second round another 7 wich is evenly devided so that iruka could just swap the names to get a normal outcome and if you take 14 as an example the students will be fighting 6 in the first round and 7 in the second. so no swapping is possible.
up until now it seems that the option is 1 or 3 since it cannot be an even
number and somebody is always left out, so wich one is it..
for that i picked the numbers 5 (1 guy left) and 7 (3 guys left)
you can only have a combat if it comes from two sides so if nr. 1 fights nr. 3 then nr. 3 must also fight nr. 1
with 5: round A =
1 vs. 2,3
2 vs. 1,4
3 vs. 1,5
4 vs. 2,5
5 vs. 4,3
with 5: round B =
1 vs. 4,5
2 vs. 3,5
3 vs. 2,4
4 vs. 1,3
5 vs. 1,2
with 7: round A
1 vs. 2,3,6
2 vs. 1,4,7
3 vs. 1,5,4
4 vs. 2,6,3
5 vs. 3,7,6
6 vs. 4,1,5
7 vs. 5,2
with 7: round B
1 vs. 4,5,7
2 vs. 3,5,6
3 vs. 2,6,7
4 vs. 1,5,7
5 vs. 1,4,2
6 vs. 2,3,7
7 vs. 1,3,4,6
As you can see with 5 you can swap numbers easy and with 7 its impossible..
so my conclusion is that there is only 1 person that will be left out.
a bit long but i hope i explained it well :)
so am i right?
Shika : Nope. The answer is a range of values. And it's incredibly simple.

(07/12) Nico : Just a random guess:
Take a class of x nins.
One ninja may fight n nins in the first round. In the second round, that nin must then fight x  (n + 1) nins to complete the condition of fighting every nin.
If the table for round 2 can be rearranged to be exactly the same as round 1, that indicates that x  (n + 1) == n , which means x = 2n + 1. This indicates that x must be odd ( x  1 cannot be equal to 2 times any number, if x is even ) , hence the remainder when divided by 4 can also therefore only be odd, i.e. 1 or 3.
I don't think it could be narrowed down further than that, but hey. Much
like Shika, I'm far too lazy to go any further.
Shika : I see some sort of reasoning here, which leads to a wrong answer.

(06/12) PCay : Ok. Since 1 person goes against 5, 4 go against 8.
*shrugs* I say 8 persons can be left out.
Shika : I guess you mean 0 then(since the remainder of any number when divided by 4 is 0,1,2,3). Please refer to previous wrong answer.

(03/12) Puizui Puizui: Dear shika, i wanted to clarify a few things
1) In the table, is the students place in groups? (eg if there are 5 students,a and b fought c . so does the table show a and b together in a box ,with arrows leading to C; or 2 seperate boxes with arrows)
2) Does"If all the students are now divided into squads of 4, what is the
possible numbers of students that will be left over. " mean that the total
number of students in iruga's class is divided by 4 (after the training)
(and thus the qn asks for the remainder ,after dividing it by 4)
Shika : With regard to 1.) it is a & b in a box labelled c, to indicate c's opponent. 2.)yes

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