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Question Number 71852 by Rio Michael last updated on 21/Oct/19

show that 5^(22)  + 17^(22)  ≡ 6 (mod 11)

$${show}\:{that}\:\mathrm{5}^{\mathrm{22}} \:+\:\mathrm{17}^{\mathrm{22}} \:\equiv\:\mathrm{6}\:\left({mod}\:\mathrm{11}\right) \\ $$

Answered by mind is power last updated on 21/Oct/19

ferma little theorem  since 11 is prime⇒5^(10) =1(11)⇒5^(20) =1^2 =1(11)  5^(22) =25(11)=3(11)  17=6(11)⇒17^(20) =6^(20) =1(11)  ⇒7^(22) =6^2 (11)=3(11)  ⇒7^(22) +5^(22) =3+3(11)=6(11)

$$\mathrm{ferma}\:\mathrm{little}\:\mathrm{theorem} \\ $$$$\mathrm{since}\:\mathrm{11}\:\mathrm{is}\:\mathrm{prime}\Rightarrow\mathrm{5}^{\mathrm{10}} =\mathrm{1}\left(\mathrm{11}\right)\Rightarrow\mathrm{5}^{\mathrm{20}} =\mathrm{1}^{\mathrm{2}} =\mathrm{1}\left(\mathrm{11}\right) \\ $$$$\mathrm{5}^{\mathrm{22}} =\mathrm{25}\left(\mathrm{11}\right)=\mathrm{3}\left(\mathrm{11}\right) \\ $$$$\mathrm{17}=\mathrm{6}\left(\mathrm{11}\right)\Rightarrow\mathrm{17}^{\mathrm{20}} =\mathrm{6}^{\mathrm{20}} =\mathrm{1}\left(\mathrm{11}\right) \\ $$$$\Rightarrow\mathrm{7}^{\mathrm{22}} =\mathrm{6}^{\mathrm{2}} \left(\mathrm{11}\right)=\mathrm{3}\left(\mathrm{11}\right) \\ $$$$\Rightarrow\mathrm{7}^{\mathrm{22}} +\mathrm{5}^{\mathrm{22}} =\mathrm{3}+\mathrm{3}\left(\mathrm{11}\right)=\mathrm{6}\left(\mathrm{11}\right) \\ $$

Commented by Rio Michael last updated on 21/Oct/19

thanks

$${thanks} \\ $$

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