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3-3-3-3-find-last-two-digits-




Question Number 174581 by Gbenga last updated on 04/Aug/22
3^3^3^3     find last two digits
3333findlasttwodigits
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
3333(mod100)φ(100)=φ(52)φ(22)=20.2=40a401(mod100),gcd(a,100)=1notethat:327=(34)633=8163333=27(mod40)327=40q+27,forsomeqZ3333=3327=340q+27=(3q)40327327=8163319633(mod100)but192=36139(mod100)1963339333=1173173(mod100)but172=28911(mod100)17311.17=18713(mod100)333313(mod100)sothelasttwodigitsare1and3
Commented by Gbenga last updated on 14/Aug/22
thankw wk
thankwwk

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