## BdMO National Secondary 2011/8

Moon
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### BdMO National Secondary 2011/8

Problem 8:
Bhaskaracharya has set up a strange study group. Any member of that group has exactly one immediate teacher (the teacher who teaches him) except for Bhaskaracharya himself, although teacher of teacher is also respected as a teacher. As the chancellor of the group, Bhaskaracharya is not taught by anybody. No two members of that group can be teachers of each other. The study group operates in a pairs where each pair consists of one member and his immediate teacher. If such a pairing is possible, is it unique? Justify your answer.
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Shapnil
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### Re: BdMO National Secondary 2011/8

I'm not sure is this correct.But I'm trying to solve it
suppose nubmers of members are n.as no two members of that group can be teachers of each other
then then all members are student (except vaskarachrya).then he's the only teacher of the group.then we can sayit is a uniqe way

*Mahi*
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### Re: BdMO National Secondary 2011/8

Use $L^AT_EX$, It makes our work a lot easier!

nafistiham
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### Re: BdMO National Secondary 2011/8

it is not unique i think.suppose, B(Bhaskaracharya) has one student, he makes a pair with him and there are $2n$ persons in the group apart from B and his student.for $n=2$ i am showing, think the $4$ persons are $a,b,c,d$
now, if the relation is cyclic, i mean,
$a$ teaches $b$
$b$ teaches $c$
$c$ teaches $d$
$d$ teaches $a$
everyone has only one immediate teachers.but, we can make both these pairings
$\binom{a}{b}\binom{c}{d}$
$\binom{d}{a}\binom{b}{c}$
the pairings are different.
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

nafistiham
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### Re: BdMO National Secondary 2011/8

[quote=
$a$ teaches Sbb$teaches Sc$
$c$ teaches Sdd$teaches Sa$
[/quote]

sorry, it should have been like this

$a$ teaches $b$
$b$ teaches $c$
$c$ teaches $d$
$d$ teaches $a$
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

*Mahi*
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### Re: BdMO National Secondary 2011/8

nafistiham wrote:
Cyclic groups are not allowed, as teacher of teacher is respected as teacher.

Use $L^AT_EX$, It makes our work a lot easier!

nafistiham
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### Re: BdMO National Secondary 2011/8

*Mahi* wrote:
nafistiham wrote:
Cyclic groups are not allowed, as teacher of teacher is respected as teacher.
I think it is unique then.because, cyclic is the only solution, which is not allowed.
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

*Mahi*
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### Re: BdMO National Secondary 2011/8

There is a really cool solution of this with graph theory. I'll post it when I have some time.

Use $L^AT_EX$, It makes our work a lot easier!

nafistiham
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### Re: BdMO National Secondary 2011/8

graph theory !! i thought it can be done by group theory
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

samiul_samin
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### Re: BdMO National Secondary 2011/8

What is the solution using graph theory?