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2017 Regional no.9 Dhaka
Posted: Sat Jun 03, 2017 5:08 pm
by Md. Rifat uddin
Re: 2017 Regional no.9 Dhaka
Posted: Mon Jun 05, 2017 2:31 pm
by aritra barua
Check the parity of the equation and the rest should be clear.
Re: 2017 Regional no.9 Dhaka
Posted: Wed Jun 07, 2017 9:05 am
by Md. Rifat uddin
So there are no solutions?? But the powers of 2 can be negative,so how do I handle that??
Re: 2017 Regional no.9 Dhaka
Posted: Wed Jun 07, 2017 2:37 pm
by ahmedittihad
No, there are solutions. You have to consider the case where $k+4-4m^2=0$ or $n^2+k-m^2=0$.
Re: 2017 Regional no.9 Dhaka
Posted: Wed Jun 07, 2017 11:04 pm
by Md. Rifat uddin
Oh !! sorry I missed that case of k+n^2-m^2 =0. SO the ans. is 17 ?? :/
Re: 2017 Regional no.9 Dhaka
Posted: Thu Jun 08, 2017 1:00 am
by Zahin Hasin Rudro
I got 16 ,how 17 ??
Re: 2017 Regional no.9 Dhaka
Posted: Thu Jun 08, 2017 3:05 pm
by Md. Rifat uddin
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.
Re: 2017 Regional no.9 Dhaka
Posted: Fri Jun 09, 2017 1:49 am
by Zahin Hasin Rudro
Oops I missed 81and 93!
Re: 2017 Regional no.9 Dhaka
Posted: Mon Jun 12, 2017 1:20 am
by Md. Rifat uddin
Btw I'm not sure id my ans. is correct :/ Am I correct??
Re: 2017 Regional no.9 Dhaka
Posted: Wed Jun 21, 2017 5:42 pm
by Abdullah Al Tanzim
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
