Dhaka Secondary 2009/11

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Secondary 2009/11

Unread post by BdMO » Fri Jan 21, 2011 6:23 pm

\[ ( 4^{502} + 2^{1004} ) ^2+(4^{502}-2^{1004} )^2-4^{2009/{2} }=2^k\]
Find $k$.

Sudip Deb
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Re: Dhaka Secondary 2009/11

Unread post by Sudip Deb » Tue Jan 25, 2011 7:26 pm

The k is 2009 .

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leonardo shawon
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Re: Dhaka Secondary 2009/11

Unread post by leonardo shawon » Tue Jan 25, 2011 9:13 pm

sudip, bro please give a full proof or at least a little hint :-)
Ibtehaz Shawon
BRAC University.

long way to go .....

Sudip Deb new
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Re: Dhaka Secondary 2009/11

Unread post by Sudip Deb new » Wed Jan 26, 2011 9:58 am

Ok I am giving some hints,
4^502 = 2^2008,
and 2 x 2^2008 = 2^2009 .
and (a+b)^2 - (a-b)^2 = 2a^2 + 2b^2 . Now try yourself .

Shifat
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Re: Dhaka Secondary 2009/11

Unread post by Shifat » Thu Aug 25, 2011 4:25 pm

darun moja pailam eta solve kore here it is:

$4^{502}= 2^{1004}$ and $4^{\frac{2009}{2}}= 2^{2009}$


so the equation is $(2.2^{1004})^2-2^{2009}= 2^k$
or $2^{2010}-2^{2009}= 2^k$
or$2^{2009}(2-1)= 2^k$
so $k=2009$

:)

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