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2023-2023-mod-13-




Question Number 201352 by cortano12 last updated on 05/Dec/23
   2023^(2023)  = ... (mod 13)
$$\:\:\:\mathrm{2023}^{\mathrm{2023}} \:=\:…\:\left(\mathrm{mod}\:\mathrm{13}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 05/Dec/23
   2023^(2023)  ≡ ... (mod 13)  2023^(2023)        ≡8^(2023) ≡x(mod 13)      [∵ 2023≡8(mod 13)]  ∵8^4 ≡1(mod 13  ∴ 8^(2023) =(8^4 )^(505) (8^3 )≡8^3 ≡5(mod 13)
$$\:\:\:\mathrm{2023}^{\mathrm{2023}} \:\equiv\:…\:\left(\mathrm{mod}\:\mathrm{13}\right) \\ $$$$\mathrm{2023}^{\mathrm{2023}} \\ $$$$\:\:\:\:\:\equiv\mathrm{8}^{\mathrm{2023}} \equiv{x}\left(\mathrm{mod}\:\mathrm{13}\right)\:\:\:\:\:\:\left[\because\:\mathrm{2023}\equiv\mathrm{8}\left(\mathrm{mod}\:\mathrm{13}\right)\right] \\ $$$$\because\mathrm{8}^{\mathrm{4}} \equiv\mathrm{1}\left({mod}\:\mathrm{13}\right. \\ $$$$\therefore\:\mathrm{8}^{\mathrm{2023}} =\left(\mathrm{8}^{\mathrm{4}} \right)^{\mathrm{505}} \left(\mathrm{8}^{\mathrm{3}} \right)\equiv\mathrm{8}^{\mathrm{3}} \equiv\mathrm{5}\left(\mathrm{mod}\:\mathrm{13}\right) \\ $$
Answered by mr W last updated on 05/Dec/23
2023^(2023)  mod 13  =(155×13+8)^(2023)  mod 13  ≡8^(2023)  mod 13  =8×(64)^(1011)  mod 13  =8×(5×13−1)^(1011)  mod 13  ≡−8 mod 13  ≡5 mod 13
$$\mathrm{2023}^{\mathrm{2023}} \:{mod}\:\mathrm{13} \\ $$$$=\left(\mathrm{155}×\mathrm{13}+\mathrm{8}\right)^{\mathrm{2023}} \:{mod}\:\mathrm{13} \\ $$$$\equiv\mathrm{8}^{\mathrm{2023}} \:{mod}\:\mathrm{13} \\ $$$$=\mathrm{8}×\left(\mathrm{64}\right)^{\mathrm{1011}} \:{mod}\:\mathrm{13} \\ $$$$=\mathrm{8}×\left(\mathrm{5}×\mathrm{13}−\mathrm{1}\right)^{\mathrm{1011}} \:{mod}\:\mathrm{13} \\ $$$$\equiv−\mathrm{8}\:{mod}\:\mathrm{13} \\ $$$$\equiv\mathrm{5}\:{mod}\:\mathrm{13} \\ $$
Answered by BaliramKumar last updated on 05/Dec/23
2023^(2023)  = x (mod13)              [φ(13)= 12]  (13×155+8)^((12×168+7))  = x(mod13)  (8)^((7))  = x(mod13)  (8^2 )^3 ×8^1  = x(mod13)  (64)^3 ×8^1  = x(mod13)  (−1)^3 ×8^1  = x(mod13)  −8 = x(mod13)  1×13−8 = x(mod13)  5 = 5(mod13)  x = 5
$$\mathrm{2023}^{\mathrm{2023}} \:=\:\mathrm{x}\:\left(\mathrm{mod13}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\phi\left(\mathrm{13}\right)=\:\mathrm{12}\right] \\ $$$$\left(\mathrm{13}×\mathrm{155}+\mathrm{8}\right)^{\left(\mathrm{12}×\mathrm{168}+\mathrm{7}\right)} \:=\:\mathrm{x}\left(\mathrm{mod13}\right) \\ $$$$\left(\mathrm{8}\right)^{\left(\mathrm{7}\right)} \:=\:\mathrm{x}\left(\mathrm{mod13}\right) \\ $$$$\left(\mathrm{8}^{\mathrm{2}} \right)^{\mathrm{3}} ×\mathrm{8}^{\mathrm{1}} \:=\:\mathrm{x}\left(\mathrm{mod13}\right) \\ $$$$\left(\mathrm{64}\right)^{\mathrm{3}} ×\mathrm{8}^{\mathrm{1}} \:=\:\mathrm{x}\left(\mathrm{mod13}\right) \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{3}} ×\mathrm{8}^{\mathrm{1}} \:=\:\mathrm{x}\left(\mathrm{mod13}\right) \\ $$$$−\mathrm{8}\:=\:\mathrm{x}\left(\mathrm{mod13}\right) \\ $$$$\mathrm{1}×\mathrm{13}−\mathrm{8}\:=\:\mathrm{x}\left(\mathrm{mod13}\right) \\ $$$$\mathrm{5}\:=\:\mathrm{5}\left(\mathrm{mod13}\right) \\ $$$$\mathrm{x}\:=\:\mathrm{5} \\ $$

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