Shikamaru's Maths Challange

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

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

(30/12) Janet Mann: The answer is that you need more information. Yes, you'll have a remainder of 0-3 but until you know how many academy students are fighting it's impossible to have a valid answer. If your example "(i.e. 1 nin can go against 5, or 3 can go against 7)" has any real bearings on the rest of the question then the number of students fighting is 10 (3 vs. 7) so the answer to the question is 2, but you've ruled that out already.

Shika : This wouldn't be my style.

-----

(28/12) Henry Ho : dunno if I'm just stupid or if I'm asking the right
question, but what's the number of students?

??????????

Shika : ... I have said what is to be said.

-----

(27/12) DuoShock : The possible number of ninjas left after assigning them to be in squads of 4 should be 1.

Thanks to your hint and a bit of trial and error. It has been proven that for remainders of 0,2,3, the 1st round of matching up the opponents could not repeat for the 2nd round. Due to the condition of not being able to change opponent for the 1st round, but this condition does not apply to the 2nd round , which means a ninja could be match up to 2 or above individuals , which could also be in different groups.

The only logical explaination for a remainder of 1 is that for a group of 5 ninjas, they could only group up to a combination of 3 and 2 ninjas matching each other. Similarly, the pattern would rearise from the 2nd round. With this, I conclude that the other numbers that left a remainder of 1 would also reapply this logic.

Did I get it right? ^^

Shika : 1 is possible, but it is not the only possibility.

-----

(20/12) Kat : well, I have a couple ideas as to what the anwser could be.

since one battle in the first round is identical to the one in the second round once relabelled, yet every student has to fight every other student in the group at least once in only the two rounds, there must be an odd number of students. In this case, the only possible remainders of students is 1 and 3.

or

relabelling could mean not switching the numbers of the students, but instead switching what he calls those students. see, if he lyed out the numbers like this:
      1    3    5    7    etc.
      2    4    6    8    etc.
then the matches would be 1-2, 3-4, 5-6, etc. for one round and 1-3-5-7-etc, 2-4-6-8-etc, for the second round. if he labelled the students in the first row that were fighting each other "odds" and the fighters in the second row "evens" for the first round, then he could just say that the odds in the first round were fighting the evens in the second round and the table would look the same, the connections of the teams would just be off.
following this logic, the number of people would be even, but there is a flaw to this logic in that not everyone would fight each other. the only way to have everyone fight each other in this is for there to be only 4 people fighting1-2 and 3-4 in the first round and 1-3 and 2-4 in the second round. in this case, the only possible number of students is 0.

if neither of these answers are correct, please tell me the reason and answer one question. in each round, is there simply one fight, or multiples. because if there are multiple fights, but the students cannot change opponents in the first round, but can in the second round, then that simplifies things for the students and complicates them for me, but it would be useful information.

please and thank you

Shika : First of all, you have 2 interpretation of the problem, in each, one is suppose to be able to know if the total no. students are odd or even. The fact is one cannot know this for sure(odd or even) from the problem, the reasoning is already wrong at this stage. And to answer your question, there can be multiple fights.

-----

(18/12) JamesBondJR703 : every one would be excluded/left over

Shika : Refer to previous postings

-----

(15/12) Cel : What I don't get is why you insist on any number divided by four.  Everyone has already said, 0, 1, 2, 3, and 4 several times, they've even said ranges of numbers.  If you're looking for a value, then how about " >1 " or " <4 ", or how about a range value like "0-4" The answer is ONE of these and you know it.

Shika : The fact is that although it is a range of numbers, the one that is absolutely impossible(given our conditions) is not in it. An example: if I say I am thinking of some even numbers from 1 to 10, and you suggest that that 1 to 10 are all values that are possible, that is obviously wrong, 3, for example would never be one of those numbers. You have to be mindful of my condition.  And this is not a guessing game.

-----

(13/12) Yello : it's been a while since anybody gave a try at solving the problem, but it looks to me that most people always give the same answers over and over.  I'll take a shot now, I hope this is the right answer, because I really want to know the answer by now, after reading all the others.

In round A, everybody is allowed to fight as many others as they want, so there can be students who fight nobody and students who fight everybody.  The problem is actually that if one student wants to fight everybody there can be nobody choosing to fight nobody.

I'd tried to figure out how the table of round B could be identical to the one of round A, and I guess the only way this is possible is... I can't explain without a chart, sorry.

Squad 1-4 (A)

Squad 5-8(B)

Squad 9-12(C)

1

2

3

4

5

6

7

8

9

10

11

12

A

12*

1

0

X

X

X

X

X

X

X

X

X

X

X

11*

2

0

X

X

X

X

X

X

X

X

X

X

10*

3

0

X

X

X

X

X

X

X

X

X

9*

4

0

X

X

X

X

X

X

X

X

B

8*

5

0

X

X

X

X

X

X

X

7*

6

0

X

X

X

X

X

X

6*

7

0

X

X

X

X

X

5*

8

0

X

X

X

X

C

4*

9

0

X

X

X

3*

10

0

X

X

2*

11

0

X

1*

12

0

12*

11*

10*

9*

8*

7*

6*

5*

4*

3*

2*

1*

So here I start (random) with 12 students (it doesn't really matter how many).  I tried to visualize everything as simple as possible, but offcourse the table can be ordered different, as long as we use the same idea behind it. The X's represent fights, so nr 12 fights everybody (he wanted to fight everybody), but also nr1 fights everybody (because everyone wanted to fight him). Or: if there is a student who fights nobody in round A, this means that he has to fight everybody in round B. And if you want an identical table, this means that there has to be someone in round A fighting everybody, from which we conclude that the option of fighting nobody in round A is impossible, for this particular situation. (and so on...) If we now re-label the students like this: nr12 becomes nr1*, nr11 becoms nr2*, ..., nr 1 becomes nr12*, we get the identical table.

(i.e. nr5 had 4 fight, so in round B he has to fight 7students; and nr8 had 7 fights in round A, so now he has to fight 4 students.  We changed there numbers, so now nr5 = nr8* and nr8= nr5*, and here we have the same as in round A, that student nr5* has 4 fights and student nr8* has 7 fights).

Ok, now students are divided in squads of 4.  To keep things simple we just take nrs1-4 together, nrs5-8 and nrs9-12.  The squads 9-12 doesn't have to fight, because at least one student of their squad has fought everybody.  Squad 1-4 neither, for the same reason.  Only left is squad 5-8, and as you can see, both available squads have already fought against at least one student in their squad.

Offcourse you can have classes with more students, but the result is the same.

This means that in Round B, in this particular situation, nobody has to fight.  My answer is: EVERYBODY will be left over.

Shika : Sorry, but you are wrong. To get a feel for the solution, take a look at those answers that I've mention are on the right track. Start from there and you will be rewarded. 

-----

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