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BdMO Online Forum • View topic - 2017 Regional no.9 Dhaka

2017 Regional no.9 Dhaka

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2017 Regional no.9 Dhaka

Post Number:#1  Unread postby Md. Rifat uddin » Sat Jun 03, 2017 5:08 pm

Anyone knows How to solve this ?? :/ :( :? :?
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Re: 2017 Regional no.9 Dhaka

Post Number:#2  Unread postby aritra barua » Mon Jun 05, 2017 2:31 pm

Check the parity of the equation and the rest should be clear.
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Re: 2017 Regional no.9 Dhaka

Post Number:#3  Unread postby Md. Rifat uddin » Wed Jun 07, 2017 9:05 am

So there are no solutions?? But the powers of 2 can be negative,so how do I handle that??
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Re: 2017 Regional no.9 Dhaka

Post Number:#4  Unread postby ahmedittihad » Wed Jun 07, 2017 2:37 pm

No, there are solutions. You have to consider the case where $k+4-4m^2=0$ or $n^2+k-m^2=0$.
Frankly, my dear, I don't give a damn.
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Re: 2017 Regional no.9 Dhaka

Post Number:#5  Unread postby Md. Rifat uddin » Wed Jun 07, 2017 11:04 pm

Oh !! sorry I missed that case of k+n^2-m^2 =0. SO the ans. is 17 ?? :/
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Re: 2017 Regional no.9 Dhaka

Post Number:#6  Unread postby Zahin Hasin Rudro » Thu Jun 08, 2017 1:00 am

I got 16 ,how 17 ??
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Re: 2017 Regional no.9 Dhaka

Post Number:#7  Unread postby Md. Rifat uddin » Thu Jun 08, 2017 3:05 pm

Actually I get 18 values .. I worked out the values of k- 15,21,27,33,35,39,51,55,57,77,99,65,69,85,87,93,95,81.
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Re: 2017 Regional no.9 Dhaka

Post Number:#8  Unread postby Zahin Hasin Rudro » Fri Jun 09, 2017 1:49 am

Oops I missed 81and 93!
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Re: 2017 Regional no.9 Dhaka

Post Number:#9  Unread postby Md. Rifat uddin » Mon Jun 12, 2017 1:20 am

Btw I'm not sure id my ans. is correct :/ Am I correct??
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Re: 2017 Regional no.9 Dhaka

Post Number:#10  Unread postby Abdullah Al Tanzim » Wed Jun 21, 2017 5:42 pm

I think the problem can be solved in this way--- 1+2^(4-3m^2-n^2)=2^(k+4-4m^2)+2^(n^2+k-m^2)
or,1+2^(4-3m^2-n^2)=2^k(2^(4-4m^2)+2^(n^2-m^2))
or,1+a/b=2^k(a+b)
or,(a+b)/b=2^k(a+b)
or,2^k=1/b
or,2^k=2^(m^2-n^2)
or,k=m^2-n^2
Then by solving this equation,I think the answer is 72
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