Chess Game!

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willpower
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Chess Game!

Unread post by willpower » Tue Nov 15, 2011 12:56 am

AoA!
A group of 61 students is divided into 3 subgroups. Two students play a game of chess (only once) if and only if they are in the same subgroup. Find a partition such that the total number of games is a multiple of 61.
What do you say?
Everybody is a genius; but if you judge a fish on its ability to climb a tree, it will live its entire life believing that it is stupid. - Albert Einstein

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nafistiham
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Re: Chess Game!

Unread post by nafistiham » Tue Nov 15, 2011 5:30 pm

i think it should have been said that,no subgroup can be an empty set.

if it is not a condition then a solution is $[0,0,61]$

hint:
suppose, the number of members of the subgroups are $x,y,z$

we know
\[x+y+z=61\]
now,
\[61k=_{2}^{x}\textrm{C}+_{2}^{y}\textrm{C}+_{2}^{z}\textrm{C}\]
\[=\frac{x!}{2!(x-2)!}+\frac{y!}{2!(y-2)!}+\frac{z!}{2!(z-2)!}\]
\[=\frac{x^2-1}{2}+\frac{y^2-1}{2}+\frac{z^2-1}{2}\]
\[=\frac{x^2+y^2+z^2-3}{2}\]
which means
\[122k+3=x^2+y^2+z^2\]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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willpower
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Re: Chess Game!

Unread post by willpower » Tue Nov 15, 2011 5:49 pm

Your answer being?
Everybody is a genius; but if you judge a fish on its ability to climb a tree, it will live its entire life believing that it is stupid. - Albert Einstein

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nafistiham
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Re: Chess Game!

Unread post by nafistiham » Wed Nov 16, 2011 6:07 pm

the next thing is trial and error. i haven't completed it yet. whenever i do, i'll post.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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willpower
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Re: Chess Game!

Unread post by willpower » Tue Nov 22, 2011 9:25 pm

You've written: \[\frac{x!}{2!(x - 2)!}= \frac{x^2 - 1}{2}\]
Shouldn't it instead be as follows?
\[\frac{x!}{2!(x - 2)!}= \frac{x^2 - x}{2}\]
Everybody is a genius; but if you judge a fish on its ability to climb a tree, it will live its entire life believing that it is stupid. - Albert Einstein

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nafistiham
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Re: Chess Game!

Unread post by nafistiham » Thu Nov 24, 2011 6:14 pm

oops. :oops: :oops: :oops:
I am really sorry about the mistake.
thanks a lot for the correction. :D
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
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willpower
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Re: Chess Game!

Unread post by willpower » Fri Nov 25, 2011 10:20 pm

There's no need to be sorry. :)
Everybody is a genius; but if you judge a fish on its ability to climb a tree, it will live its entire life believing that it is stupid. - Albert Einstein

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