2017 Regional no.9 Dhaka

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samiul_samin
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Joined: Sat Dec 09, 2017 1:32 pm

Re: 2017 Regional no.9 Dhaka

Unread post by samiul_samin » Sun Feb 17, 2019 11:40 am

Abdullah Al Tanzim wrote:
Wed Jun 21, 2017 5:42 pm
I think the problem can be solved in this way--- :arrow: $1+2^(4-3m^2-n^2)=2^(k+4-4m^2)+2^(n^2+k-m^2)$ :arrow: $1+2^(4-3m^2-n^2)=2^k(2^(4-4m^2)+2^(n^2-m^2))$
:arrow: $1+a/b=2^k(a+b)$
:arrow: $(a+b)/b=2^k(a+b)$
:arrow: $2^k=1/b$
:arrow: $2^k=2^(m^2-n^2)$
:arrow: $k=m^2-n^2$
Then by solving this equation,I think the answer is 72
I think the value of $k=20$.

This is also Mymensingh regional 2017 problem 9

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MrCriminal
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Re: 2017 Regional no.9 Dhaka

Unread post by MrCriminal » Wed Apr 07, 2021 2:20 pm

\(k=3, m=2, n=1\) works
\(k=5, m=3, n=2\) works .

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