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Question Number 201257 by BaliramKumar last updated on 02/Dec/23
Biggest prime factor of (3^(14) + 3^(13) − 12) = ?
$$\mathrm{Biggest}\:\mathrm{prime}\:\mathrm{factor}\:\mathrm{of}\:\left(\mathrm{3}^{\mathrm{14}} \:+\:\mathrm{3}^{\mathrm{13}} \:−\:\mathrm{12}\right)\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 02/Dec/23
Biggest prime factor of (3^(14) + 3^(13) − 12) = ? =3(3^(13) + 3^(12) − 4) =3(3^(13) −3 + 3^(12) −1) =3{3(3^(12) −1) + (3^(12) −1)} =3{4(3^(12) −1)} =12{(3^6 )^2 −(1)^2 } =12(3^6 −1)(3^6 +1) =12( (3^2 )^3 −1^3 )(3^2 )^3 +1^3 ) =12(3^2 −1)(3^4 +3^2 +1)(3^2 +1)(3^4 −3^2 +1) =12(8)(91)(10)(73) =12(8)(7)(13)(10)(73) 73 is the biggest prime
$$\mathrm{Biggest}\:\mathrm{prime}\:\mathrm{factor}\:\mathrm{of}\:\left(\mathrm{3}^{\mathrm{14}} \:+\:\mathrm{3}^{\mathrm{13}} \:−\:\mathrm{12}\right)\:=\:? \\ $$$$=\mathrm{3}\left(\mathrm{3}^{\mathrm{13}} \:+\:\mathrm{3}^{\mathrm{12}} \:−\:\mathrm{4}\right) \\ $$$$=\mathrm{3}\left(\mathrm{3}^{\mathrm{13}} −\mathrm{3}\:+\:\mathrm{3}^{\mathrm{12}} −\mathrm{1}\right) \\ $$$$=\mathrm{3}\left\{\mathrm{3}\left(\mathrm{3}^{\mathrm{12}} −\mathrm{1}\right)\:+\:\left(\mathrm{3}^{\mathrm{12}} −\mathrm{1}\right)\right\} \\ $$$$=\mathrm{3}\left\{\mathrm{4}\left(\mathrm{3}^{\mathrm{12}} −\mathrm{1}\right)\right\} \\ $$$$=\mathrm{12}\left\{\left(\mathrm{3}^{\mathrm{6}} \right)^{\mathrm{2}} −\left(\mathrm{1}\right)^{\mathrm{2}} \right\} \\ $$$$=\mathrm{12}\left(\mathrm{3}^{\mathrm{6}} −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{6}} +\mathrm{1}\right) \\ $$$$\left.=\mathrm{12}\left(\:\left(\mathrm{3}^{\mathrm{2}} \right)^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} \right)\left(\mathrm{3}^{\mathrm{2}} \right)^{\mathrm{3}} +\mathrm{1}^{\mathrm{3}} \right) \\ $$$$=\mathrm{12}\left(\mathrm{3}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{4}} +\mathrm{3}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{4}} −\mathrm{3}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$=\mathrm{12}\left(\mathrm{8}\right)\left(\mathrm{91}\right)\left(\mathrm{10}\right)\left(\mathrm{73}\right) \\ $$$$=\mathrm{12}\left(\mathrm{8}\right)\left(\mathrm{7}\right)\left(\mathrm{13}\right)\left(\mathrm{10}\right)\left(\mathrm{73}\right) \\ $$$$\mathrm{73}\:{is}\:{the}\:{biggest}\:{prime} \\ $$
Commented by BaliramKumar last updated on 02/Dec/23
Nice solution thanks
$$\mathrm{Nice}\:\mathrm{solution} \\ $$$$\mathrm{thanks} \\ $$
Answered by Rasheed.Sindhi last updated on 02/Dec/23
Another Way 3^(14) + 3^(13) − 12) =3^(13) (3+1)−12 =4(3^(13) −3) =4.3(3^(12) −1) =12(3^6 −1)(3^6 +1) =12{(3^2 )^3 −1^3 }{(3^2 )^3 +1^3 } =12(3^2 −1)(3^4 +3^2 +1)(3^2 +1)(3^4 −3^2 +1) =12.8.91.10.73 =12.8.7.13.10.73⇒73 is the biggest prime
$$\mathbb{A}\mathrm{nother}\:\mathbb{W}\mathrm{ay} \\ $$$$\left.\mathrm{3}^{\mathrm{14}} \:+\:\mathrm{3}^{\mathrm{13}} \:−\:\mathrm{12}\right)\: \\ $$$$=\mathrm{3}^{\mathrm{13}} \left(\mathrm{3}+\mathrm{1}\right)−\mathrm{12} \\ $$$$=\mathrm{4}\left(\mathrm{3}^{\mathrm{13}} −\mathrm{3}\right) \\ $$$$=\mathrm{4}.\mathrm{3}\left(\mathrm{3}^{\mathrm{12}} −\mathrm{1}\right) \\ $$$$=\mathrm{12}\left(\mathrm{3}^{\mathrm{6}} −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{6}} +\mathrm{1}\right) \\ $$$$=\mathrm{12}\left\{\left(\mathrm{3}^{\mathrm{2}} \right)^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} \right\}\left\{\left(\mathrm{3}^{\mathrm{2}} \right)^{\mathrm{3}} +\mathrm{1}^{\mathrm{3}} \right\} \\ $$$$=\mathrm{12}\left(\mathrm{3}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{4}} +\mathrm{3}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{4}} −\mathrm{3}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$=\mathrm{12}.\mathrm{8}.\mathrm{91}.\mathrm{10}.\mathrm{73} \\ $$$$=\mathrm{12}.\mathrm{8}.\mathrm{7}.\mathrm{13}.\mathrm{10}.\mathrm{73}\Rightarrow\mathrm{73}\:{is}\:{the}\:{biggest}\:{prime} \\ $$
Commented by Rasheed.Sindhi last updated on 03/Dec/23
An other version 3^(14) + 3^(13) − 12 =3^(14) −9 + 3^(13) − 3 =9(3^(12) −1)+3(3^(12) −1) =12(3^(12) −1) ....
$$\mathcal{A}{n}\:{other}\:{version} \\ $$$$\mathrm{3}^{\mathrm{14}} \:+\:\mathrm{3}^{\mathrm{13}} \:−\:\mathrm{12} \\ $$$$=\mathrm{3}^{\mathrm{14}} −\mathrm{9}\:+\:\mathrm{3}^{\mathrm{13}} \:−\:\mathrm{3} \\ $$$$=\mathrm{9}\left(\mathrm{3}^{\mathrm{12}} −\mathrm{1}\right)+\mathrm{3}\left(\mathrm{3}^{\mathrm{12}} −\mathrm{1}\right) \\ $$$$=\mathrm{12}\left(\mathrm{3}^{\mathrm{12}} −\mathrm{1}\right) \\ $$$$…. \\ $$

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