Round 3

For discussing Olympiad Level Number Theory problems
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Masum
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Round 3

Unread post by Masum » Thu Dec 29, 2011 10:44 pm

Deadline: $11.00$ pm tomorrow.
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One one thing is neutral in the universe, that is $0$.

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*Mahi*
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Re: Round 3

Unread post by *Mahi* » Thu Dec 29, 2011 11:24 pm

This problem set is lot easier than the earlier two! Hope everyone can do their best!
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Masum
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Re: Round 3

Unread post by Masum » Thu Dec 29, 2011 11:38 pm

*Mahi* wrote:This problem set is lot easier than the earlier two! Hope everyone can do their best!
Tomorrow is olympiad in Dhaka. So :)
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Abdul Muntakim Rafi
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Re: Round 3

Unread post by Abdul Muntakim Rafi » Fri Dec 30, 2011 1:56 am

7.There are 8 lines mutually parallel and 10 horizontal lines. Find the number
of parallelograms made by the intersections of these lines.

Didn't understand the English... :(
Bangla please...
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Phlembac Adib Hasan
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Re: Round 3

Unread post by Phlembac Adib Hasan » Fri Dec 30, 2011 8:43 am

I don't understand what prob 1 says.It's not clear what informations are given :?: .Can anyone explain it?
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Abdul Muntakim Rafi
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Re: Round 3

Unread post by Abdul Muntakim Rafi » Fri Dec 30, 2011 10:56 am

The question doesn't want you to give a numerical value,I think... Express the diameter by $AB,AC and AD$
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Nadim Ul Abrar
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Re: Round 3

Unread post by Nadim Ul Abrar » Fri Dec 30, 2011 12:40 pm

Masum wrote:5. You have 3 types of 20 books, each type has identical books. There are 5
books of the type 1, 6 of the type 7. In how many ways can you arrange them?
will it be "6 of the type 2"??
$\frac{1}{0}$

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nafistiham
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Re: Round 3

Unread post by nafistiham » Fri Dec 30, 2011 2:05 pm

Abdul Muntakim Rafi wrote:7.There are 8 lines mutually parallel and 10 horizontal lines. Find the number
of parallelograms made by the intersections of these lines.

Didn't understand the English... :(
Bangla please...

১০ টা সমান্তরাল রেখা ৮ টা সমান্তরাল রেখাকে ছেদ করলে কয়টা সামান্তরিক তৈরি হবে ?
Nadim Ul Abrar wrote:
Masum wrote:5. You have 3 types of 20 books, each type has identical books. There are 5
books of the type 1, 6 of the type 7. In how many ways can you arrange them?
will it be "6 of the type 2"??

yap. i did not understand it as well.@masum vaia
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Shihab
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Re: Round 3

Unread post by Shihab » Fri Dec 30, 2011 2:07 pm

Is it possible to solve no. 4 without calculator?
God has made the integers, all the rest is the work of man.
-Leopold Kronecker

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Nadim Ul Abrar
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Re: Round 3

Unread post by Nadim Ul Abrar » Fri Dec 30, 2011 3:16 pm

Shihab wrote:Is it possible to solve no. 4 without calculator?
:mrgreen: of course we can :mrgreen:
$\frac{1}{0}$

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