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Solve-the-system-of-congruences-2x-1-mod5-3x-2-mod7-4x-1-mod11-




Question Number 135692 by liberty last updated on 15/Mar/21
  Solve the system of congruences  2x≡1(mod5)  3x≡2(mod7)  4x≡1(mod11)
$$ \\ $$Solve the system of congruences
2x≡1(mod5)
3x≡2(mod7)
4x≡1(mod11)
Answered by floor(10²Eta[1]) last updated on 15/Mar/21
2x≡1(mod 5)⇒x≡3(mod 5)⇒x=5a+3  3(5a+3)=15a+9≡a+2≡2(mod 7)  ⇒a≡0(mod 7)⇒a=7b⇒x=35b+3  4(35b+3)=140b+12≡8b+1≡1(mod 11)  8b≡0(mod 11)⇒b≡0(mod 11)⇒b=11c  ⇒x=385c+3, c∈Z
$$\mathrm{2x}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right)\Rightarrow\mathrm{x}\equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{5}\right)\Rightarrow\mathrm{x}=\mathrm{5a}+\mathrm{3} \\ $$$$\mathrm{3}\left(\mathrm{5a}+\mathrm{3}\right)=\mathrm{15a}+\mathrm{9}\equiv\mathrm{a}+\mathrm{2}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\Rightarrow\mathrm{a}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{7}\right)\Rightarrow\mathrm{a}=\mathrm{7b}\Rightarrow\mathrm{x}=\mathrm{35b}+\mathrm{3} \\ $$$$\mathrm{4}\left(\mathrm{35b}+\mathrm{3}\right)=\mathrm{140b}+\mathrm{12}\equiv\mathrm{8b}+\mathrm{1}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{11}\right) \\ $$$$\mathrm{8b}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{11}\right)\Rightarrow\mathrm{b}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{11}\right)\Rightarrow\mathrm{b}=\mathrm{11c} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{385c}+\mathrm{3},\:\mathrm{c}\in\mathbb{Z} \\ $$
Commented by liberty last updated on 15/Mar/21
thank you
$${thank}\:{you} \\ $$

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