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Question-84941

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Question Number 84941 by Power last updated on 17/Mar/20
Commented by mr W last updated on 17/Mar/20
use method in Q84782 to find the last two digits
$${use}\:{method}\:{in}\:{Q}\mathrm{84782}\:{to}\:{find}\:{the} \\ $$$${last}\:{two}\:{digits} \\ $$
Answered by mr W last updated on 17/Mar/20
348^(67) =^2 48^(67) =48×(48^2 )^(33) =^2 48×4^(33) =48×4×(4^4 )^8 =^2 48×4×56^8 =48×4×(56^2 )^4 =^2 48×4×36^4 =48×4×(36^2 )^2 =^2 48×4×96^2 =^2 48×4×16=48×64 =^2 72 ⇒the last two digits of 348^(67) are 72 ⇒348^(67) ≡72 mod (100)
$$\mathrm{348}^{\mathrm{67}} \\ $$$$\overset{\mathrm{2}} {=}\mathrm{48}^{\mathrm{67}} =\mathrm{48}×\left(\mathrm{48}^{\mathrm{2}} \right)^{\mathrm{33}} \\ $$$$\overset{\mathrm{2}} {=}\mathrm{48}×\mathrm{4}^{\mathrm{33}} =\mathrm{48}×\mathrm{4}×\left(\mathrm{4}^{\mathrm{4}} \right)^{\mathrm{8}} \\ $$$$\overset{\mathrm{2}} {=}\mathrm{48}×\mathrm{4}×\mathrm{56}^{\mathrm{8}} =\mathrm{48}×\mathrm{4}×\left(\mathrm{56}^{\mathrm{2}} \right)^{\mathrm{4}} \\ $$$$\overset{\mathrm{2}} {=}\mathrm{48}×\mathrm{4}×\mathrm{36}^{\mathrm{4}} =\mathrm{48}×\mathrm{4}×\left(\mathrm{36}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\overset{\mathrm{2}} {=}\mathrm{48}×\mathrm{4}×\mathrm{96}^{\mathrm{2}} \\ $$$$\overset{\mathrm{2}} {=}\mathrm{48}×\mathrm{4}×\mathrm{16}=\mathrm{48}×\mathrm{64} \\ $$$$\overset{\mathrm{2}} {=}\mathrm{72} \\ $$$$\Rightarrow{the}\:{last}\:{two}\:{digits}\:{of}\:\mathrm{348}^{\mathrm{67}} \:{are}\:\mathrm{72} \\ $$$$\Rightarrow\mathrm{348}^{\mathrm{67}} \equiv\mathrm{72}\:{mod}\:\left(\mathrm{100}\right) \\ $$
Answered by mr W last updated on 18/Mar/20
what if the question is 348^(67) ≡x mod 89 348^(67) =(3×89+81)^(67) ≡81^(67) mod 89 ≡81×(73×89+64)^(33) mod 89 ≡81×64^(33) mod 89 ≡81×64×(46×89+2)^(16) mod 89 ≡81×64×2^(16) mod 89 ≡81×64×(2×89+78)^2 mod 89 ≡81×64×78^2 mod 89 ≡81×64×(68×89+32) mod 89 ≡81×64×32 mod 89 ≡(58×89+22)×32 mod 89 ≡22×32 mod 89 ≡(7×89+81) mod 89 ≡81 mod 89
$${what}\:{if}\:{the}\:{question}\:{is} \\ $$$$\mathrm{348}^{\mathrm{67}} \equiv{x}\:{mod}\:\mathrm{89} \\ $$$$ \\ $$$$\mathrm{348}^{\mathrm{67}} =\left(\mathrm{3}×\mathrm{89}+\mathrm{81}\right)^{\mathrm{67}} \\ $$$$\equiv\mathrm{81}^{\mathrm{67}} \:{mod}\:\mathrm{89} \\ $$$$\equiv\mathrm{81}×\left(\mathrm{73}×\mathrm{89}+\mathrm{64}\right)^{\mathrm{33}} \:{mod}\:\mathrm{89} \\ $$$$\equiv\mathrm{81}×\mathrm{64}^{\mathrm{33}} \:{mod}\:\mathrm{89} \\ $$$$\equiv\mathrm{81}×\mathrm{64}×\left(\mathrm{46}×\mathrm{89}+\mathrm{2}\right)^{\mathrm{16}} \:{mod}\:\mathrm{89} \\ $$$$\equiv\mathrm{81}×\mathrm{64}×\mathrm{2}^{\mathrm{16}} \:{mod}\:\mathrm{89} \\ $$$$\equiv\mathrm{81}×\mathrm{64}×\left(\mathrm{2}×\mathrm{89}+\mathrm{78}\right)^{\mathrm{2}} \:{mod}\:\mathrm{89} \\ $$$$\equiv\mathrm{81}×\mathrm{64}×\mathrm{78}^{\mathrm{2}} \:{mod}\:\mathrm{89} \\ $$$$\equiv\mathrm{81}×\mathrm{64}×\left(\mathrm{68}×\mathrm{89}+\mathrm{32}\right)\:{mod}\:\mathrm{89} \\ $$$$\equiv\mathrm{81}×\mathrm{64}×\mathrm{32}\:{mod}\:\mathrm{89} \\ $$$$\equiv\left(\mathrm{58}×\mathrm{89}+\mathrm{22}\right)×\mathrm{32}\:{mod}\:\mathrm{89} \\ $$$$\equiv\mathrm{22}×\mathrm{32}\:{mod}\:\mathrm{89} \\ $$$$\equiv\left(\mathrm{7}×\mathrm{89}+\mathrm{81}\right)\:{mod}\:\mathrm{89} \\ $$$$\equiv\mathrm{81}\:{mod}\:\mathrm{89} \\ $$

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