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3x-2-mod-5-3x-4-mod-7-3x-6-mod-11-x-




Question Number 178747 by cortano1 last updated on 21/Oct/22
   { ((3x=2 (mod 5))),((3x=4 (mod 7) )),((3x=6 (mod 11))) :}   x=?
$$\:\:\begin{cases}{\mathrm{3x}=\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{5}\right)}\\{\mathrm{3x}=\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\right)\:}\\{\mathrm{3x}=\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{11}\right)}\end{cases} \\ $$$$\:\mathrm{x}=? \\ $$
Answered by Rasheed.Sindhi last updated on 21/Oct/22
   { ((3x=2 (mod 5)....(i))),((3x=4 (mod 7)....(ii) )),((3x=6 (mod 11)....(iii))) :} ; x=?  (i)⇒3x≡2+2×5(mod 5)  ⇒x≡4(mod 5)  x=5k+4  (ii)⇒3(5k+4)≡4(mod 7)            15k+12≡4(mod 7)              15k≡−8+7(14)(mod 7)                  k≡6(mod 7)        k=7l+6  x=5(7l+6)+4=35l+34  (iii)⇒3x≡6(mod 11)               x≡2(mod 11)          35l+34≡2(mod 11)          35l≡−32(mod 11)          35l≡−32+22(11)(mod 11)              35l≡210(mod 11)                  l≡6(mod 11)         l=11m+6      x=35l+34=35(11m+6)+34       =385m+210+34=385m+244  x=244,385m+244
$$\:\:\begin{cases}{\mathrm{3x}=\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{5}\right)….\left(\mathrm{i}\right)}\\{\mathrm{3x}=\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\right)….\left(\mathrm{ii}\right)\:}\\{\mathrm{3x}=\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{11}\right)….\left(\mathrm{iii}\right)}\end{cases}\:;\:\mathrm{x}=? \\ $$$$\left({i}\right)\Rightarrow\mathrm{3}{x}\equiv\mathrm{2}+\mathrm{2}×\mathrm{5}\left({mod}\:\mathrm{5}\right) \\ $$$$\Rightarrow{x}\equiv\mathrm{4}\left({mod}\:\mathrm{5}\right) \\ $$$${x}=\mathrm{5}{k}+\mathrm{4} \\ $$$$\left({ii}\right)\Rightarrow\mathrm{3}\left(\mathrm{5}{k}+\mathrm{4}\right)\equiv\mathrm{4}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{15}{k}+\mathrm{12}\equiv\mathrm{4}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{15}{k}\equiv−\mathrm{8}+\mathrm{7}\left(\mathrm{14}\right)\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}\equiv\mathrm{6}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\:\:\:{k}=\mathrm{7}{l}+\mathrm{6} \\ $$$${x}=\mathrm{5}\left(\mathrm{7}{l}+\mathrm{6}\right)+\mathrm{4}=\mathrm{35}{l}+\mathrm{34} \\ $$$$\left({iii}\right)\Rightarrow\mathrm{3}{x}\equiv\mathrm{6}\left({mod}\:\mathrm{11}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\equiv\mathrm{2}\left({mod}\:\mathrm{11}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{35}{l}+\mathrm{34}\equiv\mathrm{2}\left({mod}\:\mathrm{11}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{35}{l}\equiv−\mathrm{32}\left({mod}\:\mathrm{11}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{35}{l}\equiv−\mathrm{32}+\mathrm{22}\left(\mathrm{11}\right)\left({mod}\:\mathrm{11}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{35}{l}\equiv\mathrm{210}\left({mod}\:\mathrm{11}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{l}\equiv\mathrm{6}\left({mod}\:\mathrm{11}\right) \\ $$$$\:\:\:\:\:\:\:{l}=\mathrm{11}{m}+\mathrm{6} \\ $$$$\:\:\:\:{x}=\mathrm{35}{l}+\mathrm{34}=\mathrm{35}\left(\mathrm{11}{m}+\mathrm{6}\right)+\mathrm{34} \\ $$$$\:\:\:\:\:=\mathrm{385}{m}+\mathrm{210}+\mathrm{34}=\mathrm{385}{m}+\mathrm{244} \\ $$$${x}=\mathrm{244},\mathrm{385}{m}+\mathrm{244} \\ $$
Commented by Tawa11 last updated on 21/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by cortano1 last updated on 21/Oct/22
nice. thanks sir
$$\mathrm{nice}.\:\mathrm{thanks}\:\mathrm{sir} \\ $$

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