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Question Number 174581 by Gbenga last updated on 04/Aug/22

3^3^3^3 find last two digits

$$\mathrm{3}^{\mathrm{3}^{\mathrm{3}^{\mathrm{3}} } } \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{last}}\:\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{digits}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by floor(10²Eta[1]) last updated on 05/Aug/22

3^3^3^3 (mod100) ϕ(100)=ϕ(5^2 )ϕ(2^2 )=20.2=40 ⇒a^(40) ≡1(mod100), gcd(a,100)=1 note that: 3^(27) =(3^4 )^6 3^3 =81^6 3^3 ≡3^3 =27(mod40) ⇒3^(27) =40q+27, for some q∈Z 3^3^3^3 =3^3^(27) =3^(40q+27) =(3^q )^(40) 3^(27) ≡3^(27) =81^6 3^3 ≡19^6 3^3 (mod100) but 19^2 =361≡−39(mod100) ⇒19^6 3^3 ≡39^3 3^3 =117^3 ≡17^3 (mod100) but 17^2 =289≡−11(mod100) ⇒17^3 ≡−11.17=−187≡13(mod100) ⇒3^3^3^3 ≡13(mod100) so the last two digits are 1 and 3

$$\mathrm{3}^{\mathrm{3}^{\mathrm{3}^{\mathrm{3}} } } \left(\mathrm{mod100}\right) \\ $$$$\varphi\left(\mathrm{100}\right)=\varphi\left(\mathrm{5}^{\mathrm{2}} \right)\varphi\left(\mathrm{2}^{\mathrm{2}} \right)=\mathrm{20}.\mathrm{2}=\mathrm{40} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{40}} \equiv\mathrm{1}\left(\mathrm{mod100}\right),\:\mathrm{gcd}\left(\mathrm{a},\mathrm{100}\right)=\mathrm{1} \\ $$$$\mathrm{note}\:\mathrm{that}: \\ $$$$\mathrm{3}^{\mathrm{27}} =\left(\mathrm{3}^{\mathrm{4}} \right)^{\mathrm{6}} \mathrm{3}^{\mathrm{3}} =\mathrm{81}^{\mathrm{6}} \mathrm{3}^{\mathrm{3}} \equiv\mathrm{3}^{\mathrm{3}} =\mathrm{27}\left(\mathrm{mod40}\right) \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{27}} =\mathrm{40q}+\mathrm{27},\:\mathrm{for}\:\mathrm{some}\:\mathrm{q}\in\mathbb{Z} \\ $$$$\mathrm{3}^{\mathrm{3}^{\mathrm{3}^{\mathrm{3}} } } =\mathrm{3}^{\mathrm{3}^{\mathrm{27}} } =\mathrm{3}^{\mathrm{40q}+\mathrm{27}} =\left(\mathrm{3}^{\mathrm{q}} \right)^{\mathrm{40}} \mathrm{3}^{\mathrm{27}} \equiv\mathrm{3}^{\mathrm{27}} \\ $$$$=\mathrm{81}^{\mathrm{6}} \mathrm{3}^{\mathrm{3}} \equiv\mathrm{19}^{\mathrm{6}} \mathrm{3}^{\mathrm{3}} \left(\mathrm{mod100}\right) \\ $$$$\mathrm{but}\:\mathrm{19}^{\mathrm{2}} =\mathrm{361}\equiv−\mathrm{39}\left(\mathrm{mod100}\right) \\ $$$$\Rightarrow\mathrm{19}^{\mathrm{6}} \mathrm{3}^{\mathrm{3}} \equiv\mathrm{39}^{\mathrm{3}} \mathrm{3}^{\mathrm{3}} =\mathrm{117}^{\mathrm{3}} \equiv\mathrm{17}^{\mathrm{3}} \left(\mathrm{mod100}\right) \\ $$$$\mathrm{but}\:\mathrm{17}^{\mathrm{2}} =\mathrm{289}\equiv−\mathrm{11}\left(\mathrm{mod100}\right) \\ $$$$\Rightarrow\mathrm{17}^{\mathrm{3}} \equiv−\mathrm{11}.\mathrm{17}=−\mathrm{187}\equiv\mathrm{13}\left(\mathrm{mod100}\right) \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{3}^{\mathrm{3}^{\mathrm{3}} } } \equiv\mathrm{13}\left(\mathrm{mod100}\right) \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{1}\:\mathrm{and}\:\mathrm{3} \\ $$

Commented by Gbenga last updated on 14/Aug/22

thankw wk

$$\boldsymbol{\mathrm{thankw}}\:\boldsymbol{\mathrm{wk}} \\ $$

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