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

(18/11) Dr Picker : Okay I'm back with hopefully a solution that you find suitable.

There are x (total) number of ninjas who are fighting with each other.  If n=1 ninjas and x/4=(# of squads) then 4n=1 squad as you said "If all the students are now divided into squads of 4..."  x/4=4n then 16n=x or
1/16x=n. x/4=16n/4.  since n=1 there are 16 squads total.  x=16.  With the amount of squads established we can now move on to a domain restriction on y which is the possible amount of students leftover. 16*4=64, the total number of students.  1/4 of the students can NOT fight 1/4 of the number of students as they will fight themselves which is a highly improbable thing that could occur.  They also can NOT fight the other 3/4 as they would be fighting in their same squad.  this leaves y with a range of 0<y<4. (When I say the "less than" I mean "less than or equal to" but yahoo does not allow me to have the underline.)

Now, in Round B, if youw ere to set up the round as they can not fight the same opponents, they would only be able to fight with 64 total in round A, but not everyone fought everyone, half and half (too lazy to
do the probablity).  1/4 fights 1/2 and another 1/4 and then in round B the other 1/4 fights the half as that would not be fighting the same opponents and the 1st 1/4 will fight amongst themselves.  By a simple
relabelling of x (being first 1/4), y (1/2) and z (second 1/4) to z, y and x you can get the identical tables that Iruka got.

Thus, in conclusion, I hope I have achieved an answer befitting for you.  Thank you.

Shika : I think this is not the correct aproach. Think simpler.

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(18/11) Alejandro Gonzalez : Okay, hear me out here. Two rounds, students bunched in groups of four. Assuming that all groups are filled with four students, and there is at least one group,there will always be an even number of potential fights. Assuming this is the groups:

a     b
c     d

Each student seven potential fights. What I mean is: a vs b, a vs c, a vs d, a vs bcd, a vs bc, a vs bd, a vs cd. It is that similar pattern for student in the group of four. Thus, the equation for the number of potential fights is this:

P*(P*4)*7=R

P is the number of groups (multiplied by four to get number of people) Seven represents the number of possibilities per person R is the result: the number of potential fights.

The result of this equation represents the number of potential fights in round A, since the exact number of students in his class is unknown.

In round B, since they cannot fight the same students as they did in round A, and must fight all the students they didn't in round A, logically, the number of potential fights would be the potential fights of the previous round minus the potential fights of the previous round divided by the number fights that did occur. What I mean is this:

F=R-(R/28)       or               F=R-(R/7)

F is the number of fights that did occur in round B. R is the potential fights from the previous round R divided by twenty-eight is the  absolute least  number of fights that could have occurred (assuming, of course, that no one sat at the sidelines and did nothing) in round A, and thus could not be repeated. R divided by seven is the absolute most number of fights that could have occurred, and thus could not be repeated.

The result, F, is the number of fights that did occur in round B, based on whether the number of fights that occured was the maximum number of fights, or the minimum number of fights, or if it was somewhere in between.

Now, the fights that occured in round A, minus the fights that occured in round B, gives you the number of fights that occured in neither round, which is essentially the possible fights left over. Basically this:

R-F=X

R is from equation one, the number of fights occuring in round A. F is from equation two, the number of fights occuring in round B. X is the fights that occured in neither round.

Taking the square root of X will warrant the number of people left over. Since the final value of the square root of what you got for X will always been a positive number at least one or greater, the number of people left over is any number between one and positive infinity. Since we do not know the precise number of students in Iruka's class, all we have is the range.

The range will always be at least one, because the square root of the smallest possible result of the final equation will be one, and it can extend as far as infinity, because you do not have any limiting values that could potentially set an upper limit on the range.

Wow, that took a lot of typing!

Shika : Frankly speaking it takes more brains to read your thesis than in solving the actual problem, and probably much more in cooking it up. I think the solution is very easily within your reach if you would give it more though. 

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