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Question-189417




Question Number 189417 by sonukgindia last updated on 16/Mar/23
Answered by mehdee42 last updated on 16/Mar/23
(2^(3141) )^2 +(2^x )^2 +(2^(1618) )^2 =(2^a +2^b )^2 =(2^a )^2 +(2^b )^2 +2^(a+b+1)   1)if a=3141 & b=1618 & a+b+1=2x⇒x=2380  2)if a=3141 &b=x&a+b+1=3236⇒x=94  3)a=1618&b=x&a+b+1=6282⇒x=4663  ⇒s=7137
$$\left(\mathrm{2}^{\mathrm{3141}} \right)^{\mathrm{2}} +\left(\mathrm{2}^{{x}} \right)^{\mathrm{2}} +\left(\mathrm{2}^{\mathrm{1618}} \right)^{\mathrm{2}} =\left(\mathrm{2}^{{a}} +\mathrm{2}^{{b}} \right)^{\mathrm{2}} =\left(\mathrm{2}^{{a}} \right)^{\mathrm{2}} +\left(\mathrm{2}^{{b}} \right)^{\mathrm{2}} +\mathrm{2}^{{a}+{b}+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right){if}\:{a}=\mathrm{3141}\:\&\:{b}=\mathrm{1618}\:\&\:{a}+{b}+\mathrm{1}=\mathrm{2}{x}\Rightarrow{x}=\mathrm{2380} \\ $$$$\left.\mathrm{2}\right){if}\:{a}=\mathrm{3141}\:\&{b}={x\&a}+{b}+\mathrm{1}=\mathrm{3236}\Rightarrow{x}=\mathrm{94} \\ $$$$\left.\mathrm{3}\right){a}=\mathrm{1618\&}{b}={x\&a}+{b}+\mathrm{1}=\mathrm{6282}\Rightarrow{x}=\mathrm{4663} \\ $$$$\Rightarrow{s}=\mathrm{7137} \\ $$

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