solve-the-system-of-linear-congruences-x-2-mod-3-x-4-mod-5-x-7-mod-9-x-11-mod-13-using-the-Brute-force-method-

Question Number 71838 by Rio Michael last updated on 20/Oct/19

$${solve}\:{the}\:{system}\:{of}\:{linear}\:{congruences}\: \\ $$$$\:{x}\:\equiv\:\mathrm{2}\:\left({mod}\:\mathrm{3}\right) \\ $$$${x}\:\equiv\:\mathrm{4}\left({mod}\:\mathrm{5}\right) \\ $$$${x}\:\equiv\:\mathrm{7}\:\left({mod}\:\mathrm{9}\right) \\ $$$${x}\equiv\:\mathrm{11}\left(\:{mod}\:\mathrm{13}\right) \\ $$$${using}\:{the}\:{Brute}\:{force}\:{method} \\ $$

Commented by mind is power last updated on 20/Oct/19

⇔ x≡4(5) x≡7(9) x≡11(13) x≡4(5),x≡7(9)⇔x≡34 (45) our system⇔ { ((x≡34(45))),((x≡11(13))) :}⇔x≡ 349(585) solution x≡349(585)

$$\Leftrightarrow \\ $$$$\mathrm{x}\equiv\mathrm{4}\left(\mathrm{5}\right) \\ $$$$\mathrm{x}\equiv\mathrm{7}\left(\mathrm{9}\right) \\ $$$$\mathrm{x}\equiv\mathrm{11}\left(\mathrm{13}\right) \\ $$$$\mathrm{x}\equiv\mathrm{4}\left(\mathrm{5}\right),\mathrm{x}\equiv\mathrm{7}\left(\mathrm{9}\right)\Leftrightarrow\mathrm{x}\equiv\mathrm{34}\:\:\left(\mathrm{45}\right) \\ $$$$\mathrm{our}\:\mathrm{system}\Leftrightarrow\begin{cases}{\mathrm{x}\equiv\mathrm{34}\left(\mathrm{45}\right)}\\{\mathrm{x}\equiv\mathrm{11}\left(\mathrm{13}\right)}\end{cases}\Leftrightarrow\mathrm{x}\equiv\:\mathrm{349}\left(\mathrm{585}\right) \\ $$$$\mathrm{solution}\:\mathrm{x}\equiv\mathrm{349}\left(\mathrm{585}\right) \\ $$

Commented by mr W last updated on 21/Oct/19

x ≡ 2 (mod 3) and x ≡ 7 (mod 9) are contradictory! x=9k+7=3(3k+2)+1=3m+1≠3m+2

$${x}\:\equiv\:\mathrm{2}\:\left({mod}\:\mathrm{3}\right)\:{and}\: \\ $$$${x}\:\equiv\:\mathrm{7}\:\left({mod}\:\mathrm{9}\right)\:{are}\:{contradictory}! \\ $$$${x}=\mathrm{9}{k}+\mathrm{7}=\mathrm{3}\left(\mathrm{3}{k}+\mathrm{2}\right)+\mathrm{1}=\mathrm{3}{m}+\mathrm{1}\neq\mathrm{3}{m}+\mathrm{2} \\ $$

Commented by mind is power last updated on 21/Oct/19

$$\mathrm{yeah}\: \\ $$

Answered by mr W last updated on 21/Oct/19

x=13a+11=9b+7 9b−13a=4 ⇒b=13c−1, a=9c+1 x=9(13c−1)+7=117c−2=5d+4 117c−5d=6 ⇒d=117e+69, c=5e+3 x=5(117e+69)+4=585e+349=^(!) 3f+2 3(f−195e)=347 ⇒impossible x ≡ 2 (mod 3) is contradiction! solution is x=585n+349

$${x}=\mathrm{13}{a}+\mathrm{11}=\mathrm{9}{b}+\mathrm{7} \\ $$$$\mathrm{9}{b}−\mathrm{13}{a}=\mathrm{4} \\ $$$$\Rightarrow{b}=\mathrm{13}{c}−\mathrm{1},\:{a}=\mathrm{9}{c}+\mathrm{1} \\ $$$${x}=\mathrm{9}\left(\mathrm{13}{c}−\mathrm{1}\right)+\mathrm{7}=\mathrm{117}{c}−\mathrm{2}=\mathrm{5}{d}+\mathrm{4} \\ $$$$\mathrm{117}{c}−\mathrm{5}{d}=\mathrm{6} \\ $$$$\Rightarrow{d}=\mathrm{117}{e}+\mathrm{69},\:{c}=\mathrm{5}{e}+\mathrm{3} \\ $$$${x}=\mathrm{5}\left(\mathrm{117}{e}+\mathrm{69}\right)+\mathrm{4}=\mathrm{585}{e}+\mathrm{349}\overset{!} {=}\mathrm{3}{f}+\mathrm{2} \\ $$$$\mathrm{3}\left({f}−\mathrm{195}{e}\right)=\mathrm{347}\:\Rightarrow{impossible} \\ $$$${x}\:\equiv\:\mathrm{2}\:\left({mod}\:\mathrm{3}\right)\:{is}\:{contradiction}! \\ $$$$ \\ $$$${solution}\:{is}\:{x}=\mathrm{585}{n}+\mathrm{349} \\ $$

Commented by Rio Michael last updated on 21/Oct/19

$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$

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