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 | ||
$$\Leftrightarrow \\ $$$$\mathrm{x}\equiv\mathrm{4}\left(\mathrm{5}\right) \\ $$$$\mathrm{x}\equiv\mathrm{7}\left(\mathrm{9}\right) \\ $$$$\mathrm{x}\equiv\mathrm{11}\left(\mathrm{13}\right) \\ $$$$\mathrm{x}\equiv\mathrm{4}\left(\mathrm{5}\right),\mathrm{x}\equiv\mathrm{7}\left(\mathrm{9}\right)\Leftrightarrow\mathrm{x}\equiv\mathrm{34}\:\:\left(\mathrm{45}\right) \\ $$$$\mathrm{our}\:\mathrm{system}\Leftrightarrow\begin{cases}{\mathrm{x}\equiv\mathrm{34}\left(\mathrm{45}\right)}\\{\mathrm{x}\equiv\mathrm{11}\left(\mathrm{13}\right)}\end{cases}\Leftrightarrow\mathrm{x}\equiv\:\mathrm{349}\left(\mathrm{585}\right) \\ $$$$\mathrm{solution}\:\mathrm{x}\equiv\mathrm{349}\left(\mathrm{585}\right) \\ $$ | ||
Commented by mr W last updated on 21/Oct/19 | ||
$${x}\:\equiv\:\mathrm{2}\:\left({mod}\:\mathrm{3}\right)\:{and}\: \\ $$$${x}\:\equiv\:\mathrm{7}\:\left({mod}\:\mathrm{9}\right)\:{are}\:{contradictory}! \\ $$$${x}=\mathrm{9}{k}+\mathrm{7}=\mathrm{3}\left(\mathrm{3}{k}+\mathrm{2}\right)+\mathrm{1}=\mathrm{3}{m}+\mathrm{1}\neq\mathrm{3}{m}+\mathrm{2} \\ $$ | ||
Commented by mind is power last updated on 21/Oct/19 | ||
$$\mathrm{yeah}\: \\ $$ | ||
Answered by mr W last updated on 21/Oct/19 | ||
$${x}=\mathrm{13}{a}+\mathrm{11}=\mathrm{9}{b}+\mathrm{7} \\ $$$$\mathrm{9}{b}−\mathrm{13}{a}=\mathrm{4} \\ $$$$\Rightarrow{b}=\mathrm{13}{c}−\mathrm{1},\:{a}=\mathrm{9}{c}+\mathrm{1} \\ $$$${x}=\mathrm{9}\left(\mathrm{13}{c}−\mathrm{1}\right)+\mathrm{7}=\mathrm{117}{c}−\mathrm{2}=\mathrm{5}{d}+\mathrm{4} \\ $$$$\mathrm{117}{c}−\mathrm{5}{d}=\mathrm{6} \\ $$$$\Rightarrow{d}=\mathrm{117}{e}+\mathrm{69},\:{c}=\mathrm{5}{e}+\mathrm{3} \\ $$$${x}=\mathrm{5}\left(\mathrm{117}{e}+\mathrm{69}\right)+\mathrm{4}=\mathrm{585}{e}+\mathrm{349}\overset{!} {=}\mathrm{3}{f}+\mathrm{2} \\ $$$$\mathrm{3}\left({f}−\mathrm{195}{e}\right)=\mathrm{347}\:\Rightarrow{impossible} \\ $$$${x}\:\equiv\:\mathrm{2}\:\left({mod}\:\mathrm{3}\right)\:{is}\:{contradiction}! \\ $$$$ \\ $$$${solution}\:{is}\:{x}=\mathrm{585}{n}+\mathrm{349} \\ $$ | ||
Commented by Rio Michael last updated on 21/Oct/19 | ||
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$ | ||