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 ----------------------------------------

(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 (x-1).

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

If the number of opponents (x-1) 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 e-mails 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(N-1) and so on...
a team of 2 can fight with all others(N-2) and so

a team A(1 person) fight with N-1 > left out
a team A(1 person) fight with N-2 > left out
a team A(1 person) fight with N-3 > left out
a team A(1 person) fight with N-4 > left out

a team A(2 persons) fight with N-2 > left out
a team A(2 persons) fight with N-3 > left out
a team A(2 persons) fight with N-4 > left out

a team A(3 persons) fight with N-3 > left out
a team A(3 persons) fight with N-4 > left out

a team A(4 persons) fight with N-4 > 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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