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Question Number 127940 by bramlexs22 last updated on 03/Jan/21

{ ((3x=1 (mod 4))),((4x=3 (mod 5) )),((5x=7 (mod 11))) :}

$$\:\begin{cases}{\mathrm{3x}=\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{4}\right)}\\{\mathrm{4x}=\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{5}\right)\:}\\{\mathrm{5x}=\mathrm{7}\:\left(\mathrm{mod}\:\mathrm{11}\right)}\end{cases} \\ $$

Answered by floor(10²Eta[1]) last updated on 03/Jan/21

3x≡1(mod 4)⇒x≡3(mod 4) x=4a+3 4(4a+3)≡3(mod 5) ⇒a≡1(mod 5)⇒a=5b+1 ⇒x=4(5b+1)+3=20b+7 5(20b+7)≡7(mod 11) ⇒b≡5(mod 11) b=11c+5⇒x=20(11c+5)+7 x=220c+107, c∈Z

$$\mathrm{3x}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{4}\right)\Rightarrow\mathrm{x}\equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{4}\right) \\ $$$$\mathrm{x}=\mathrm{4a}+\mathrm{3} \\ $$$$\mathrm{4}\left(\mathrm{4a}+\mathrm{3}\right)\equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{a}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right)\Rightarrow\mathrm{a}=\mathrm{5b}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{4}\left(\mathrm{5b}+\mathrm{1}\right)+\mathrm{3}=\mathrm{20b}+\mathrm{7} \\ $$$$\mathrm{5}\left(\mathrm{20b}+\mathrm{7}\right)\equiv\mathrm{7}\left(\mathrm{mod}\:\mathrm{11}\right) \\ $$$$\Rightarrow\mathrm{b}\equiv\mathrm{5}\left(\mathrm{mod}\:\mathrm{11}\right) \\ $$$$\mathrm{b}=\mathrm{11c}+\mathrm{5}\Rightarrow\mathrm{x}=\mathrm{20}\left(\mathrm{11c}+\mathrm{5}\right)+\mathrm{7} \\ $$$$\mathrm{x}=\mathrm{220c}+\mathrm{107},\:\mathrm{c}\in\mathbb{Z} \\ $$

Answered by liberty last updated on 04/Jan/21

{ ((x=3 (mod 4)...(i))),((x=2 (mod 5)...(ii))),((x=8 (mod 11)...(iii))) :} for(i) ⇒ 55a = 3 (mod 4) −a = 3 (mod 4) ; a=−3 (mod 4) for(ii)⇒44b=2 (mod 5) 4b=2 (mod 5) ; b=3(mod 5) for(iii)⇒20c=8 (mod 11) −2c = 8(mod 11);c=−4 (mod 11) general solution ∴ 55(−3)+44(3)+20(−4)+220k ; k∈Z i.e 220k −113 or 220k +107 ; k∈Z

$$\:\begin{cases}{\mathrm{x}=\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{4}\right)...\left(\mathrm{i}\right)}\\{\mathrm{x}=\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{5}\right)...\left(\mathrm{ii}\right)}\\{\mathrm{x}=\mathrm{8}\:\left(\mathrm{mod}\:\mathrm{11}\right)...\left(\mathrm{iii}\right)}\end{cases} \\ $$$$\mathrm{for}\left(\mathrm{i}\right)\:\Rightarrow\:\mathrm{55a}\:=\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{4}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{a}\:=\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{4}\right)\:;\:\mathrm{a}=−\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{4}\right) \\ $$$$\mathrm{for}\left(\mathrm{ii}\right)\Rightarrow\mathrm{44b}=\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{5}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4b}=\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{5}\right)\:;\:\mathrm{b}=\mathrm{3}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{for}\left(\mathrm{iii}\right)\Rightarrow\mathrm{20c}=\mathrm{8}\:\left(\mathrm{mod}\:\mathrm{11}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2c}\:=\:\mathrm{8}\left(\mathrm{mod}\:\mathrm{11}\right);\mathrm{c}=−\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{11}\right) \\ $$$$\mathrm{general}\:\mathrm{solution}\: \\ $$$$\:\therefore\:\mathrm{55}\left(−\mathrm{3}\right)+\mathrm{44}\left(\mathrm{3}\right)+\mathrm{20}\left(−\mathrm{4}\right)+\mathrm{220k}\:;\:\mathrm{k}\in\mathbb{Z} \\ $$$$\:\mathrm{i}.\mathrm{e}\:\mathrm{220k}\:−\mathrm{113}\:\mathrm{or}\:\mathrm{220k}\:+\mathrm{107}\:;\:\mathrm{k}\in\mathbb{Z} \\ $$

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