Question Number 133761 by liberty last updated on 24/Feb/21
![Solve the linear congruence 19x ≡ 4 (mod 141 )](https://www.tinkutara.com/question/Q133761.png)
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{linear}\:\mathrm{congruence}\: \\ $$$$\mathrm{19}{x}\:\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{141}\:\right) \\ $$
Answered by bobhans last updated on 24/Feb/21
![Using Euclidean algorithm 141 = 7×19+8 19 = 2×8 +3 8 = 2×3 +2 3 = 1×2 + 1 now we get 1 = 3−1×2 1=3−1×(8−2×3) 1 = 3×3−1×8 1=3×(19−2×8)−1×8 1= 3×19−7×8 1=3×19−7(141−7×19) 1=52×19−7×141 This show that 52 is an inverse of 19 (mod 141) Then 19x≡4 (mod 141) ⇒52×19x ≡ 4×52 (mod 141) ⇒(1+7×141)x ≡ 208 (mod 141) ∴ x≡ 67 (mod 141)](https://www.tinkutara.com/question/Q133762.png)
$${Using}\:{Euclidean}\:{algorithm}\: \\ $$$$\:\mathrm{141}\:=\:\mathrm{7}×\mathrm{19}+\mathrm{8} \\ $$$$\:\:\:\mathrm{19}\:=\:\mathrm{2}×\mathrm{8}\:+\mathrm{3} \\ $$$$\:\:\:\:\:\mathrm{8}\:=\:\mathrm{2}×\mathrm{3}\:+\mathrm{2}\: \\ $$$$\:\:\:\:\:\mathrm{3}\:=\:\mathrm{1}×\mathrm{2}\:+\:\mathrm{1} \\ $$$${now}\:{we}\:{get}\:\mathrm{1}\:=\:\mathrm{3}−\mathrm{1}×\mathrm{2}\: \\ $$$$\mathrm{1}=\mathrm{3}−\mathrm{1}×\left(\mathrm{8}−\mathrm{2}×\mathrm{3}\right) \\ $$$$\mathrm{1}\:=\:\mathrm{3}×\mathrm{3}−\mathrm{1}×\mathrm{8} \\ $$$$\mathrm{1}=\mathrm{3}×\left(\mathrm{19}−\mathrm{2}×\mathrm{8}\right)−\mathrm{1}×\mathrm{8} \\ $$$$\mathrm{1}=\:\mathrm{3}×\mathrm{19}−\mathrm{7}×\mathrm{8} \\ $$$$\mathrm{1}=\mathrm{3}×\mathrm{19}−\mathrm{7}\left(\mathrm{141}−\mathrm{7}×\mathrm{19}\right) \\ $$$$\mathrm{1}=\mathrm{52}×\mathrm{19}−\mathrm{7}×\mathrm{141} \\ $$$${This}\:{show}\:{that}\:\mathrm{52}\:{is}\:{an}\:{inverse}\:{of} \\ $$$$\mathrm{19}\:\left({mod}\:\mathrm{141}\right) \\ $$$${Then}\:\mathrm{19}{x}\equiv\mathrm{4}\:\left({mod}\:\mathrm{141}\right) \\ $$$$\Rightarrow\mathrm{52}×\mathrm{19}{x}\:\equiv\:\mathrm{4}×\mathrm{52}\:\left({mod}\:\mathrm{141}\right) \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{7}×\mathrm{141}\right){x}\:\equiv\:\mathrm{208}\:\left({mod}\:\mathrm{141}\right) \\ $$$$\therefore\:{x}\equiv\:\mathrm{67}\:\left({mod}\:\mathrm{141}\right) \\ $$