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Question Number 133412 by EDWIN88 last updated on 22/Feb/21
What are the last two digits of 2^(222) −1 ?
$$\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{2}^{\mathrm{222}} −\mathrm{1}\:? \\ $$
Answered by liberty last updated on 22/Feb/21
2^(10) =1024≡24 (mod 100) 2^(20) ≡24^2 =576≡ 76≡−24(mod 100) Hence 2^(30) ≡−24^2 ≡−24 (mod 100) 2^(40) ≡−24 (mod 100) and so on Because 222 = 220+2 ; 2^(222) ≡−24×2^2 ≡−96 ≡ 4 (mod 100) therefore 2^(222) −1≡3 (mod 100) The last two digits of 2^(222) −2 are 03
$$\mathrm{2}^{\mathrm{10}} =\mathrm{1024}\equiv\mathrm{24}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{20}} \equiv\mathrm{24}^{\mathrm{2}} =\mathrm{576}\equiv\:\mathrm{76}\equiv−\mathrm{24}\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{Hence}\:\mathrm{2}^{\mathrm{30}} \equiv−\mathrm{24}^{\mathrm{2}} \equiv−\mathrm{24}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{40}} \equiv−\mathrm{24}\:\left(\mathrm{mod}\:\mathrm{100}\right)\:\mathrm{and}\:\mathrm{so}\:\mathrm{on} \\ $$$$\mathrm{Because}\:\mathrm{222}\:=\:\mathrm{220}+\mathrm{2}\:;\:\mathrm{2}^{\mathrm{222}} \equiv−\mathrm{24}×\mathrm{2}^{\mathrm{2}} \\ $$$$\equiv−\mathrm{96}\:\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{therefore}\:\mathrm{2}^{\mathrm{222}} −\mathrm{1}\equiv\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{100}\right)\: \\ $$$$\mathrm{The}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{2}^{\mathrm{222}} −\mathrm{2}\:\mathrm{are}\:\mathrm{03} \\ $$

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