2017 Regional no.9 Dhaka

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samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

Re: 2017 Regional no.9 Dhaka

Abdullah Al Tanzim wrote:
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)\$ \$1+2^(4-3m^2-n^2)=2^k(2^(4-4m^2)+2^(n^2-m^2))\$
\$1+a/b=2^k(a+b)\$
\$(a+b)/b=2^k(a+b)\$
\$2^k=1/b\$
\$2^k=2^(m^2-n^2)\$
\$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

MrCriminal
Posts: 8
Joined: Wed Mar 24, 2021 6:43 pm

Re: 2017 Regional no.9 Dhaka

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