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

(01/12) Daniel Juckett : It doesn't seem like anybody has answered in a couple of months but if no one has solved it yet.

I think the answer is 1 or 0

Reasoning:

(I'm using an example to help explain the idea)

The table of who battles who should look something like this

1 -> 2, 3, 4, 5

2 -> 3, 4, 5

3 -> 4, 5

4 -> 5

5 -> no one

For the second round you just switch 1 with 5, 2 with 4, and leave 3 the same. If there are an even number of people then there is no number the stays the same.

This table then indicates that no one is left out.  However the students are in groups of 4.  This means that there could be one group that could have 1 4 people in it. If you just switch the groups, which would mean switching the students' numbers within the group, there could be somebody left out of the second round because there is no number to switch with.

For example: (using the table above)

Group 1 has students 1, 2, 3, 4

Group 5 has students 5, 6, 7

When these groups switch somebody from group one will have the same number and not complete the training thus being left out.

The answer is 1 or 0 because group 1 compared to group 5 can have no more than one more person in it.  If group 5 has only 2 people then Iruka could make group 1 and 5 have 3 people each thus leaving nobody out.  You can arrange the groups so that there is either one odd person or no odd person based on if there is an even or odd number of people in the class.

Therefore the answer has to be 0 or 1.

Shika : Could you give an explanation not depending on specific examples? Since we want absolute certainty.

-----

(01/12) EMMIBOO3 : All right, got my bro to help out with this one. We tried to break down the problem as best we can and solve it from there. All notes are included...but it's still probably wrong.

#1: A student must have fought against the three other students that he/she is put into a squad with in round A (unless these are left over)

#2: The relabeling of the students must depend on the outcome and setup of the first round.

#3: A student must fight at least (n-1) times, unless this is one of the possibilities for being left over. It is not possible to fight against yourself.

#4: Unknowns-
The actual number of students
The number of students who fought together in round A
The number of students who fought together as opponents in round A
The numbers that the students were given at the beginning

#5: Assumptions (Might not be safe)-
The answer is lower than four students because 4 is the needed number and you won't have any left over with four. (0, 1, 2, 3)
Each student does have an actual number
In round A, one or more students could have chosen not to fight against anyone
If the students were divided into squads of four in round B, and the table is supposed to be identical to the one in round A, then there was at least one group that fought as a four-person group in the first round

Round A:

Let's have a four person cell be F (and ignore any other possible 4 person squads)

If the opposing students to F had fewer numbers that F, then it would be

F:3
F:2
F:1
or F:0 (this being possible if no one fought in the first round at all)

F=4 students

Round B:

To make the chart identical, the students are relabeled.

F has students 1, 2, 3, and 4
The others have 5, 6, and 7 since any higher numbers would just confuse us further right now.

In round A, there were 4 possible battles (that matter in the final tally)-

1 5
2 6
3 7
4
______________
1 5
2 6
3
4
______________
1 5
2
3
4
______________
1
2
3
4

The only one that can be arranged to match the pervious chart is-

1 5                   1(6)           5(3)
2 6              making it        2(7)      6(4)
3                   3(1)
4                   4(2)

Since 3 and 4 have already faced 5 and 6, these are the two that are left over because, even though they complete the graph, they have already fought all opponents and are not needed. If there are any other students, then they were already put into four man squads and are not important. My answer is 2 leftover students.

P.S.: Bro says that if you want to get specific, the 3 and 4 are considered left over as well, because they don't have a 4 student squad. If this is true, then the answer is 4 and not two. Take whichever one you want.

Shika : The answer is a range of values but not all of 0,1,2,3. The argument is actually quite simple and can be solved mentally.

-----

(01/12) William George : The answer is 0 (zero). The question said : " If ALL the students are now divided into squads of 4, what is the possible numbers of students that will be left over."  so if all the students were divided in to squads then there wouldn't be anyone left. ALL the students are now in squads. 

Shika : As can be understood from the phase "squads of 4"; means 4 in a squad. If it is not, then it is the remainder.

-----

(29/11) William George : If everyone has to fight everyone once, regardless of how you split them, no one would be left over. So, left over nins are ZERO.

Shika : Do you mean to say that the total no. of students is a multiple of 4? How does this follow from your argument?

-----

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