\[ ( 4^{502} + 2^{1004} ) ^2+(4^{502}-2^{1004} )^2-4^{2009/{2} }=2^k\]
Find $k$.
Dhaka Secondary 2009/11
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Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
- leonardo shawon
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Re: Dhaka Secondary 2009/11
sudip, bro please give a full proof or at least a little hint
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....
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Re: Dhaka Secondary 2009/11
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 .
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 .
Re: Dhaka Secondary 2009/11
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 $k=2009$
$4^{502}= 2^{1004}$ 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 $k=2009$