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Question Number 92211 by jagoll last updated on 05/May/20

4x = 2 (mod 3 )

$$\mathrm{4x}\:=\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{3}\:\right)\: \\ $$

Commented by jagoll last updated on 05/May/20

x = 3k−1   x = 2,5,8,11,14,...

$$\mathrm{x}\:=\:\mathrm{3k}−\mathrm{1}\: \\ $$$$\mathrm{x}\:=\:\mathrm{2},\mathrm{5},\mathrm{8},\mathrm{11},\mathrm{14},...\: \\ $$

Answered by Rio Michael last updated on 07/May/20

 4x ≡ 2( mod 3)  here we still have one solution  since 4 and 3 are relatively prime.   4 = 1(3)+ 1   4 = 3(1) + 0  ⇒ 1 = 4−3      1 +3 = 4  multiplicative inverse is 1.   1 ×4x ≡ 1×2(mod 3)         x ≡ 2 (mod 3) should be our only solution

$$\:\mathrm{4}{x}\:\equiv\:\mathrm{2}\left(\:\mathrm{mod}\:\mathrm{3}\right) \\ $$$$\mathrm{here}\:\mathrm{we}\:\mathrm{still}\:\mathrm{have}\:\mathrm{one}\:\mathrm{solution} \\ $$$$\mathrm{since}\:\mathrm{4}\:\mathrm{and}\:\mathrm{3}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime}. \\ $$$$\:\mathrm{4}\:=\:\mathrm{1}\left(\mathrm{3}\right)+\:\mathrm{1} \\ $$$$\:\mathrm{4}\:=\:\mathrm{3}\left(\mathrm{1}\right)\:+\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{1}\:=\:\mathrm{4}−\mathrm{3} \\ $$$$\:\:\:\:\mathrm{1}\:+\mathrm{3}\:=\:\mathrm{4} \\ $$$$\mathrm{multiplicative}\:\mathrm{inverse}\:\mathrm{is}\:\mathrm{1}. \\ $$$$\:\mathrm{1}\:×\mathrm{4}{x}\:\equiv\:\mathrm{1}×\mathrm{2}\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:{x}\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{3}\right)\:\mathrm{should}\:\mathrm{be}\:\mathrm{our}\:\mathrm{only}\:\mathrm{solution} \\ $$$$ \\ $$

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