Question-151265

Question Number 151265 by liberty last updated on 19/Aug/21

Commented by Tawa11 last updated on 19/Aug/21

$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$

Commented by liberty last updated on 19/Aug/21

5x ≡ 2 (mod 3)⇒x=1+3a 3x ≡ 4 (mod 7) ⇒9a+3≡4 (mod 7) ⇒2a ≡ 1 (mod 7) ⇒a = 4+7b 3x ≡ 7 (mod 8)⇒3(1+3(4+7b))≡ 7(mod 8) ⇒3(13+21b)≡ 7 (mod 8) ⇒63b+39 ≡ 7 (mod 8) ⇒7b+7 ≡ 7 (mod 8) ⇒7b ≡ 0 (mod 8) ⇒b ≡ 0 (mod 8) ⇒b = 8c then x = 1+3a = 1+3(4+7b) x=21b+13 = 21(8c)+13 x=168c +13 the smallest positive integer x = 13 .

$$\mathrm{5x}\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{3}\right)\Rightarrow\mathrm{x}=\mathrm{1}+\mathrm{3a} \\ $$$$\mathrm{3x}\:\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\right)\:\Rightarrow\mathrm{9a}+\mathrm{3}\equiv\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\Rightarrow\mathrm{2a}\:\equiv\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{7}\right)\:\Rightarrow\mathrm{a}\:=\:\mathrm{4}+\mathrm{7b} \\ $$$$\mathrm{3x}\:\equiv\:\mathrm{7}\:\left(\mathrm{mod}\:\mathrm{8}\right)\Rightarrow\mathrm{3}\left(\mathrm{1}+\mathrm{3}\left(\mathrm{4}+\mathrm{7b}\right)\right)\equiv\:\mathrm{7}\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{13}+\mathrm{21b}\right)\equiv\:\mathrm{7}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\Rightarrow\mathrm{63b}+\mathrm{39}\:\equiv\:\mathrm{7}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\Rightarrow\mathrm{7b}+\mathrm{7}\:\equiv\:\mathrm{7}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\Rightarrow\mathrm{7b}\:\equiv\:\mathrm{0}\:\left(\mathrm{mod}\:\mathrm{8}\right)\:\Rightarrow\mathrm{b}\:\equiv\:\mathrm{0}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\Rightarrow\mathrm{b}\:=\:\mathrm{8c}\: \\ $$$$\mathrm{then}\:\mathrm{x}\:=\:\mathrm{1}+\mathrm{3a}\:=\:\mathrm{1}+\mathrm{3}\left(\mathrm{4}+\mathrm{7b}\right) \\ $$$$\mathrm{x}=\mathrm{21b}+\mathrm{13}\:=\:\mathrm{21}\left(\mathrm{8c}\right)+\mathrm{13} \\ $$$$\mathrm{x}=\mathrm{168c}\:+\mathrm{13}\: \\ $$$$\mathrm{the}\:\mathrm{smallest}\:\mathrm{positive}\:\mathrm{integer}\: \\ $$$$\mathrm{x}\:=\:\mathrm{13}\:.\: \\ $$

Commented by otchereabdullai@gmail.com last updated on 22/Aug/21

$$\mathrm{nice} \\ $$

Answered by Rasheed.Sindhi last updated on 19/Aug/21

5x≡2+3=5(mod 3) x≡1(mod 3)..................(i) 3x≡4(mod 7 3x≡4+7×2=18(mod 7 x≡6(mod7)....................(ii) 3x≡7(mod 8) 3x≡7+8=15(mod 8) x≡5(mod8)....................(iii) (i)⇒x=1,4,7,10,13,16,19,.... (ii)⇒6,13,20,27,.... (iii)5,13,21,29,... x=13

$$\mathrm{5}{x}\equiv\mathrm{2}+\mathrm{3}=\mathrm{5}\left({mod}\:\mathrm{3}\right) \\ $$$${x}\equiv\mathrm{1}\left({mod}\:\mathrm{3}\right)………………\left({i}\right) \\ $$$$\mathrm{3}{x}\equiv\mathrm{4}\left({mod}\:\mathrm{7}\right. \\ $$$$\mathrm{3}{x}\equiv\mathrm{4}+\mathrm{7}×\mathrm{2}=\mathrm{18}\left({mod}\:\mathrm{7}\right. \\ $$$${x}\equiv\mathrm{6}\left({mod}\mathrm{7}\right)………………..\left({ii}\right) \\ $$$$\mathrm{3}{x}\equiv\mathrm{7}\left({mod}\:\mathrm{8}\right) \\ $$$$\mathrm{3}{x}\equiv\mathrm{7}+\mathrm{8}=\mathrm{15}\left({mod}\:\mathrm{8}\right) \\ $$$${x}\equiv\mathrm{5}\left({mod}\mathrm{8}\right)………………..\left({iii}\right) \\ $$$$\left({i}\right)\Rightarrow{x}=\mathrm{1},\mathrm{4},\mathrm{7},\mathrm{10},\mathrm{13},\mathrm{16},\mathrm{19},…. \\ $$$$\left({ii}\right)\Rightarrow\mathrm{6},\mathrm{13},\mathrm{20},\mathrm{27},…. \\ $$$$\left({iii}\right)\mathrm{5},\mathrm{13},\mathrm{21},\mathrm{29},… \\ $$$${x}=\mathrm{13} \\ $$

Commented by liberty last updated on 19/Aug/21

$$\mathrm{okay}.\:\mathrm{thanks} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply