Shikamaru's Maths Challange

Read the IQ Challenge here. To take up the challenge, email Shikamaru your answer at

Other answers to this question : Main, Pg 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2


Challenge #04 - Training at the Academy

Your Answers ----------------------------------------

(18/12) Tim Watson : The possible numbers of students that will be left over is not a specific number but rather a category of numbers. That category is even numbers. I found that the number of students isn't relevant; therefore, there is no specific answer.

Shika : Yes it is a range of numbers but are more specific and would not contain those that would be impossible, given our conditions.


(17/12) Japple : In round 2, it's impossible for each student to spar with the nins that they haven't sparred with in the first round, and at the same time not spar with their opponents in the previous round. In order for that to happen, there would need to be a 3rd round. You are right that the second round can be identical to the first round, but some of the nin would still be sparring with previous opponents. Therefore, there is no  specific answer, but any integer can be the possible answer (except #s 0-3).

Shika : Please refer to previous postings on similar wrong answers.


(14/12) Sakura Kinimoto : Erm...I'm guessing it's 4. Cos I saw this  (i.e. 1 nin can go against 5, or 3 can go against 7) and in there, the difference between 1 and 5, 3 and 7 is 4. And since you never state that there was a certain number of people, it is (I think) safe to use the numbers you left in the question to get the people left over. So I guess it'll be 4 people or 1 squad left over. Er...Is that correct? Please send me an email to tell me if it's correct. Thanks!

Shika : Never rely on the numbers I give, they are just there to facilitate explanation.


(13/12) Uzumaki Toshiro : As stated in the rules of engagement, a student may engage as many opponents in Round A as they see fit.

And with Round B, the students must engage all opponents they did not fight in Round A.

Keeping these two rules in mind, eventually a student must face ALL opponents they did not previously fight.

As shown in one of the examples, a single student may fight as many opponents as they choose. This shows that while Squad-based combat is encouraged, it is not required.


The assumptions are based upon the above interpretation of the rules of engagement:

   1. The possibility of combat between members of the same squad.

   2. The phrase 'nins cannot change opponents' is interpreted as
      that a student may not stop fighting a selected opponent until a
      clear victory or defeat.

   3. A student may continue fighting after a defeat.


This set of rules ensures that if a squad of students successfully defeats all opponents, then in Round B the members of the squad will have to face each other. This ensures their participation in Round B.

The only possible way for there to be inactive students in Round B is for the students in question to defeat ALL opponents in Round A.


Let the letter 'n' represent the number of students participating in the combat. As a student cannot fight himself, the number of opponents
possible must be (n-1).

The possibility of challenging all opponents, in addition to the possibility of Assumption 3, allows a student to be left out of Round B.

Following the above reasoning, it is possible for ALL students to challenge ALL opponents. This would mean that no one would go on to Round B.

Of course, between the above two possibilities lies the opportunity for any number of students to attempt to take on all opponents.

Unlikely that it could ever happen, the above three examples show the
following range of non-combatants in Round B:

(0 to n)

Perhaps I have assumed too much and gotten too little sleep but this is the answer I have found.

Shika : Nope. Get some sleep then do a little meditation.


(12/12) Chanteru-chan : I made up the team numbers but the outcome will remain the same.

Key: Team ? (ooo)= 3 man team
    7 nin (ooo ooo o) = 7 ninjas

We have this in Round 1:

Team 7(ooo)against 7 nin(ooo ooo o)
Team 10(ooo)against 7 nin(ooo ooo o)
Team 9(ooo)against 7 nin(ooo ooo o)
Team 8(ooo)against 7 nin(ooo ooo o)

Here, we take out 3 nins from each of the 7 nin. This is because we're assuming that each team that consistes of 3 people, took out 3 people from each 7 nin. Now the 7 nin look like this:

7 nin(ooo ooo o) change to 4 nin(ooo o)

Since they are now grouped as 4 nin, they are already paired as squads. I shall call them by Team 11,12,13,and 14 numerically equal to Team 7,10,9,and 8.

This is how teams 7,9,10,and 8 get grouped into squads...

Team 7 gets a member from Team 9, leaving Team 9 with 2 people.

Team 10 gets a member from Team 8, leaving Team 8 with 2 people as well...

Team 9 and Team 8, because they are both left with 2 members, join forces, equaling to a squad... Team 15.

Now, for the battling:

Team 7 fought Team 11 before, therefore cannot battle them again. Also, having a member from Team 9, they cannot battle Team 13. They shall battle Team 12.

Team 10 fought Team 12 before and Team 7 is already facing them, therefore cannot battle them again anyway. Also, having a member from Team 8, they cannot battle Team 14. They shall battle Team 11.

Team 15, because they are combined from Team 9 and 8, cannot fight against Team 13 or 14 because they faced against them before. Therefore, they cannot battle anyone...

This leaves Team 15 out...

Team 13 and Team 14 fight against each other.

This leaves 4 people out, which are Team 15's squad...

If I get this wrong,I WILL try again! ^_^

I have another method but I'll try this first...

Shika : Maybe you should show me your other method.


(12/12) Lariengalasse : either 1 or 3 students will be left.
start with x number of students. Unless a student has watched Fight Club one too many times, he cannot fight himself. Therefore, the number of possible opponents is X-1. because regardles of how many students he fights in round A, he fights all the rest in round B, A +B = x-1
Because Iruka can use the same table twice, we know that A = B
so now we have the information we need
A + B = X-1
A = B

because any number added to itself is even, X - 1 is even, so X is odd.
Because X - 1 is not divisible by 4, the remainder will always be either 1 or 3

Shika : Although your answer is wrong, I must congradulate you for being one of the people who got the closest to the solution. Keep trying.


(12/12) Lariengalasse : Without knowing the original number of students, the problem is impossible to solve. The obvious answer would be that there would be either 1, 2, or 3 students left over, if there were 4 or more, then 4 of those could be made into a team. However, the fact that students can't fight the same opponents in round B as they did in round A negates that, because there could be any number of students. The closest you can get to a real answer is no more than 1/2 of the original number of students, because no matter how many students fought with any number of others, at least half of them did not fight at least 1/2 of the students.

Shika : This is not an impossible problem, just wait till you see the solution.


(11/12) Ash: The answer is 4 students because all of the students have been divided into groups of 4 (this is not an assumption, it is stated in the question). So if all of the students are divided into groups of 4 and
the number of groups are even then there won't be anyone left out to
fight.(zero is not an option because you already ruled that out!) But,
if the number of groups are odd, there will be one group that is left
out of the fight. So, the answer is 4.

Shika : I am afraid you do not have a correct understanding of the question. Please ask a friend for opinion or check to see other answers.


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