Round 3

For discussing Olympiad Level Number Theory problems
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Nadim Ul Abrar
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Re: Round 3

Unread post by Nadim Ul Abrar » Sat Dec 31, 2011 1:28 pm

Phlembac Adib Hasan wrote: We need two pairs of parallel lines to make a parallelogram.So the answer is $ {}^ {10} C_2 {}^8 C_2 =1260 $.
vai sol ta shundor hoise : :D
$\frac{1}{0}$

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Labib
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Re: Round 3

Unread post by Labib » Sat Dec 31, 2011 2:19 pm

ফিবোনাচ্চি ক্রমের যোগফল নির্নয় করার সাধারন সূত্র আছে কি?
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Abdul Muntakim Rafi
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Re: Round 3

Unread post by Abdul Muntakim Rafi » Sat Dec 31, 2011 4:13 pm

Adib Hasan: Thank you very much for a nice solution...
Nadim ul Abrar: How did u solve 7? Please explain...
Labib: Yes there are... Nadim wrote the one needed... The sum of the first n Fibonacci numbers with odd indices is what Nadim wrote...
And guys share how u proved 8 has no solution...
Man himself is the master of his fate...

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sm.joty
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Re: Round 3

Unread post by sm.joty » Sat Dec 31, 2011 4:25 pm

ফিবোনাচ্চি ক্রমের যোগফল নির্নয় করার সাধারন সূত্র আছে কি?
অবশ্যই আছে। Wikipedia তে দেখ। খুব সোজা জিনিস। :)

$f_{2012}=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{2012}-(\frac{1-\sqrt{5}}{2})^{2012}]$
I don't think its possible to have the solution without using any calculator by your way.
I have also use almost same method. So I'm also fail to do it (without calculator.)
And without calculator it's quite boring. 8-)
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

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Nadim Ul Abrar
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Re: Round 3

Unread post by Nadim Ul Abrar » Sat Dec 31, 2011 6:02 pm

@ rafi :8. what about $p=q=r$

7, My explanation must require the figure .... :(
$\frac{1}{0}$

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*Mahi*
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Re: Round 3

Unread post by *Mahi* » Sat Dec 31, 2011 8:10 pm

Labib wrote:ফিবোনাচ্চি ক্রমের যোগফল নির্নয় করার সাধারন সূত্র আছে কি?
\[\sum^n_{k=0} f_k= f_{2n+1}-1\]
\[\sum^n_{k=0} f_{2k}= f_{2n+1}\]
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*Mahi*
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Re: Round 3

Unread post by *Mahi* » Sat Dec 31, 2011 8:12 pm

Easy solution of $\boxed 8$
Let $p > q \geq r$
Then $p > \frac {q+r} 2$
But the highest divisor of $n$ may be $\frac n2$
Do the rest :)
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Nadim Ul Abrar
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Re: Round 3

Unread post by Nadim Ul Abrar » Sat Dec 31, 2011 8:28 pm

*Mahi* wrote: But the highest divisor of $n$ may be $\frac n2$
Do the rest :)
*Mahi* vai isn't n is a divisor of n
$\frac{1}{0}$

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*Mahi*
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Re: Round 3

Unread post by *Mahi* » Sat Dec 31, 2011 8:30 pm

Nadim Ul Abrar wrote:
*Mahi* wrote: But the highest divisor of $n$ may be $\frac n2$
Do the rest :)
*Mahi* vai isn't n is a divisor of n
Yes, but as the case $p=q+r$ is not possible so we can strike that out.
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Phlembac Adib Hasan
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Re: Round 3

Unread post by Phlembac Adib Hasan » Sat Dec 31, 2011 8:33 pm

I don't think its possible to have the solution without using any calculator by your way.
I have also use almost same method. So I'm also fail to do it (without calculator.)
And without calculator it's quite boring. 8-)
$f_{2012} $ has 420 digits in decimal expression.[From wolframalpha]So I don't think you can find it using ordinary scientific calculators or may be by your PC! 8-)
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