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

$$solvethesystemoflinearcongruences$$$$x\equiv 2\left(mod3\right)$$$$x\equiv 4\left(mod5\right)$$$$x\equiv 7\left(mod9\right)$$$$x\equiv 11\left(mod13\right)$$$$usingtheBruteforcemethod$$

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

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

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

$$x\equiv 2\left(mod3\right)and$$$$x\equiv 7\left(mod9\right)arecontradictory!$$$$x=9k+7=3(3k+2)+1=3m+1\ne 3m+2$$

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

$$\mathrm{yeah}$$

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

$$x=13a+11=9b+7$$$$9b-13a=4$$$$\Rightarrow b=13c-1,a=9c+1$$$$x=9(13c-1)+7=117c-2=5d+4$$$$117c-5d=6$$$$\Rightarrow d=117e+69,c=5e+3$$$$x=5(117e+69)+4=585e+349\stackrel{!}{=}3f+2$$$$3(f-195e)=347\Rightarrow impossible$$$$x\equiv 2\left(mod3\right)iscontradiction!$$$$$$$$solutionisx=585n+349$$

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

$$thankyousomuchsir$$

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