Question #250615

Solve the following:

(a) 1231001 (mod 101)

(b) 17123 (mod 13)



1
Expert's answer
2022-01-25T16:32:18-0500

Euler's Theorem states, For relatively prime integers a and n

aϕ(n)1(modn)a^{\phi(n)}\equiv 1(\mod n).

(a)Here, a = 123, n = 101 and (123,101) = 1. Using Euler’s Theorem,123ϕ(101)1(mod101)1231001(mod101)1231000(123100)101101(mod101)1231001123(mod101)22(mod101)(a) \text{Here, a = 123, n = 101 and (123,101) = 1. Using Euler's Theorem,}\\123^{\phi(101)}\equiv 1(\mod 101)\\ 123^{100}\equiv 1(\mod 101)\\ 123^{1000}\equiv (123^{100})^{10} \equiv 1^{10}\equiv1(\mod 101)\\ 123^{1001}\equiv 123(\mod 101)\equiv 22(\mod 101) \\


(b)Here, a = 17, n = 13 and (17,13) = 1. Using Euler’s Theorem,17ϕ(13)1(mod13)17121(mod13)17120(1712)10110(mod13)1(mod13)1712317120173(mod13)17123173(mod13)43(mod13)  [Since 174mod13]1712312(mod13)(b) \text{Here, a = 17, n = 13 and (17,13) = 1. Using Euler's Theorem,}\\ 17^{\phi(13)} \equiv 1(\mod 13)\\ 17^{12} \equiv 1(\mod 13)\\ 17^{120} \equiv (17^{12})^{10}\equiv 1^{10}(\mod 13)\equiv 1(\mod13)\\ 17^{123} \equiv 17^{120}\cdot 17^{3}(\mod 13)\\ 17^{123} \equiv 17^{3}(\mod 13)\equiv 4^3(\mod 13) ~~[\text{Since} ~17\equiv4\mod13]\\ 17^{123} \equiv 12(\mod 13)


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