
Challenge #05  Training at the Academy unsolved
Iruka devised a way to train the ninja academy students in the art of mass combat. These mock battles are divided into 2 rounds. Round A and Round B.
Round A : The students are allowed to fight against anyone and as many opponents they choose. (i.e. 1 nin can go against 5, or 3 can go against 7) However, the nins cannot change opponents.
Round B : The students must not spar with the opponents encounted in the previous round and must spar with all those they didn't fight in the first round.
This is to ensure that the weakest student will have a chance to fight the strongest of the lot and that the ninjawannabes will be trained in fighting as a part of a group or against many. Each student is tagged numerically and Iruka draws up a table that lists the opponents of each student.
Question: In one particular battle, Iruka discovered that the table he gets in the 2nd round could be made identical to that in the 1st round simply by relabelling the students. If all the students are now divided into squads of 4, what is the possible numbers of students that will be left over.
Confused by the results, Iruka seeks Shika for help. And Shika being lazy decides to post this as a challenge.
*Answer this Maths question correctly and enter our Hall of Fame! Correct answers includes the correct reasoning. The answer will be revealed when someone finally gets the answer correct. You can try more than once.
Your Answers 
(06/01) Inna Manalo
: I really can't explain what I did and how i did it... but the answer that i got is 7
Shika : but couldn't you further divide 7 by 4

(06/01) Muy Rei : umm...i will just say the answer...i think it's 2 Shika : Wrong even so.

(06/12) Klave : i have come up with a short answer.i decided to
dwell on when can the list be identical,disregarding
all the numbers of the group.i'll go directly if there
is only one leftover nin...if iruka were to relabel
the students like this...
(im going for the smallest variable with 2 groups and
1 extra besides same goes for 1 group and 1 extra)
group 1=a,b,c,d
group 2=e,f,g,h (letters or number it doesn't matter)
+1 nin named Naruto labeled letter N...
then if Naruto decides to fight together with group 1
so battle 1 goes like this:
a,b,c,d,N vs. e,f,g,h
then at round 2 concentrating on Naruto's list then he
would fight a,b,c,d so
at round 2
N vs. a,b,c,d...
e vs f,g,h\
f vs e,g,h \ten long tedious battles
g vs e,f,h /
h vs e,f,g/
so just looking at Naruto...he would have to spar with
four opponents
then relabeling them would give you
N vs e,f,g,h
so with the point of having the same number of
opponents in the next round after being relabeled.
if there were 1 or 3 students left over they would get
a possibility of having the same number of opponents
in the next round...
e.g
round 1
a,b,c vs d,e,f,g (lowest variable)
round 2 concentrating on a
a vs b,c
so you get a different number of opponents in the
second
round
though there are certainly times that 1 or 3 cannot
make up an identical list depending on who they fought
on the first round...but 2 and 0 being left behind
could never give you an identical list
I hope this is right...
Shika :
There's already a 1 or 3, and I did mention that it's
wrong.

(05/01) Ryan Chong
: heres my idea of how it goes.
basically the numbesr can go up to as much has
infinite in this kind of problem
Round 1: is saying that anyone can battle each other
with as much as they want against as much as they want
giving it an infinite amount but lets say theres 48
students total. but they cant change there opponents
so they stick with the same ones.
round 1.

4v4 5v3 7v6
2v6 1v2 2v6 48 students total

it says that the students vs different opponents and
not the same ones. thus giving the weaker ones a
chance to fight the stronger ones so im just gonna
change the team numbers so they fight different amount
of the other teams
round 2.
4v1 4v6 5v6
7v2 2v6 2v3
the numbers can go up to infinite. its technically
saying like 20v16
4 4
16 12 thus making it having 28 students
out of that battle left over dividing that 4 7 other
groups. and so on. the amount of students is no
concern but its dependent on how many participate in
one round affecting how much the others fight in the
second round. it can have so many possibilities. i see
the range between hundreds. my other example of the idea is that since it depends
on how many students participate in one round
affecting the other. it was stated that they cannot
fight the same student in the other team. but still
dividing the team by squasd of 4. i dont know the
exact range of how much but it goes beyond of what i
wanna try to know.
if it was 40 vs 40. and technically they have to fight
against their own teammates. cuz its to fight against
and with that matters and to arrange the teams in
fours and make sure they dont fight each other.
the range im thinking is probably 1thousands area.
its based on how many fight against one to go against
the other.
lets say its 10 students
4v2 2v2 then it'll be 4v4 2 left on it can go on and
on and on.
Shika :
Wait till you see the solution........

(05/01) Heero :
To be fair to (27/11) Cel, the problem is worded ambiguously in such a way as to cause confusion, that said...
First, in order for Iruka to get such a state (the same tables trick) each student must basically fight against "half" the class in one round and the other half in the second round. If the name swap is uneven the tables won't be identical (Round one will have one "team" (altho they're not all allies) larger than the other and thus Round Two won't have enough slots for the Round One members)
With that known, the total students must be an EVEN number the TOTAL ninja must be divisible by 2 not any given set of opponents. Thus Iruka could have EITHER 0 or 2 remaining students after breaking them into teams of 4.
Example class seen below:
Naruto=A, Sasuke=B, Sakura=C, Ino=D, Shikamaru=E, Chouji=F
Start with the knowledge that none can fight THEMSELF (Naruto maybe, but point stands as he's an exception)
xABCDEF
Ax+++++
B+x++++
C++x+++
D+++x++
E++++x+
F+++++x
Now, if Team Naruto (one way or another) Fights team Shikamaru in Round 1, then they will fight each other in Round 2 and vice versa. Since the inverse is true, you can simply swap the names and the table still stands as valid. ie. Team Shikamaru fights Team Naruto in Round one and fights itself in round 2. The table can be expanded infinitely as long as at least one member as added to each side in the first round (and thus becomes an opponent in the second.
Shika : Wrong. Next!

(05/01) Maringa (aka vampyre_smiles) ^,,^
: One little thing: If you have a freeforall, you
can't have matches (or rounds) of 2 vs 5, 3 vs 7, or x
vs y nins where neither x nor y equal 0 or 1, because
those would imply teams and you don't have teams in a
freeforall.
But then, you also never mentioned if it could be 1 vs
3 vs 7 or something like that. If so, I'll send in an
actual answer soon.
Shika :
1 vs 3 vs 7 is nonexplanatory, it could mean all
combination, which is a unnecessary restriction, since
I could have 1vs3 and 1vs7 at the same time without
3vs7

(04/01) Twilight
:
Shika, I know that this will seem like a simple approach and is probably off
the beaten path, but It seems a bit logical, I guess.
Since you are asking for a range of values I shall make a guess that the
answer lies anywhere in between the maximum Number of students and 2.
meaning the the range of values is 2the number of students in Iruka's
class. My reasoning here is actually quite simple.
Dertermination of maximum amount of students left out: All of the students
can be left out if everyone fights everyone else. Since the nins can decide
how many others they wish to fight it is logical to say that the upper
limitation of those being left out is the number of students in the class.
Therefore there can be no more fighting in round B because everyone has
fought everyone else.
Determination of the minimum number of students left out: For this step I
went on an assumption of 30 students being in Irukas class, which I deduced
from episode 3 of the Naruto episodes. Given that 10 teams of 3 were created
from the graduates, there are 30 students. Also in the episodes it was
stated that Naruto was the only student that failed to graduate. Therefore
30 is a safe bet. Anyway, If you increase the number of students per group
by one that would leave you with a remainder of 2 people. 30/4=7.5.
Therefore I can conclude that the minimum value would be at least 2.
Now there is the possibility, however unlikly,that no one would fight anyone
else meaning that none could be left out making the minimum 0, but I doubt
this to be so, because there is always at least one person like Naruto in
classes.
My
Answer:{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}
or simply {230}.
A Troublesome problem if I might say so, but I thought I'd try just the
same.
Shika : I can see that you miss it by a mile. You must therefore solve this problem. It could change your life! Maybe your mind could undergo a paradigm shift or something......

(02/01)
Han : HI!!
i read this Challenge #05  Training at the Academy, and i wanna give it a
try, if i may.
i believe, due to the information provided and following the table (1 nin
can go against 5, or 3 can go against 7) that there are 13 people minimum,
since they have to change opponents in each round.
if divided by 4, the answer would be 1.
what do you reckon?
Shika : Numerically your answer is wrong. And why should there be a minimum of 13 people? I hope you didn't take my example too seriously, they are there to help you understand and do not represent an actual fact about the problem.

Main, Pg 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2

Previous Challenges
Challenge #04  "Pieces of Donuts"
Challenge #03  "Count the Ninjas"
Challenge #02  "Find the fake Shuriken"
Challenge #01  "Find the Safe Scroll"

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