Menu Close

Question-110281




Question Number 110281 by 675480065 last updated on 28/Aug/20
Answered by Rio Michael last updated on 28/Aug/20
 3x ≡ 1 (mod 5)    (i)   5x ≡ 2 (mod 7)     (ii)  from (i)  x ≡ 2 (mod 5)  (iii)  from (ii) x ≡ 6 (mod 7)  (iv)   from (iii)  x = 5k + 2 ,  k ∈ N (v)  into (iv) ⇒ 5k + 2 ≡ 6 (mod 7)    ⇒ 5k ≡ 4 (mod 7 ) ⇔  k ≡ 5( mod 7) ⇒ k = 7t + 5 , t ∈ N  hence x = 5(7t + 5) + 2 = 35t + 27  x = 35t + 27 ,
$$\:\mathrm{3}{x}\:\equiv\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{5}\right)\:\:\:\:\left({i}\right) \\ $$$$\:\mathrm{5}{x}\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{7}\right)\:\:\:\:\:\left({ii}\right) \\ $$$$\mathrm{from}\:\left({i}\right)\:\:{x}\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{5}\right)\:\:\left({iii}\right) \\ $$$$\mathrm{from}\:\left({ii}\right)\:{x}\:\equiv\:\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{7}\right)\:\:\left({iv}\right) \\ $$$$\:\mathrm{from}\:\left({iii}\right)\:\:{x}\:=\:\mathrm{5}{k}\:+\:\mathrm{2}\:,\:\:{k}\:\in\:\mathbb{N}\:\left({v}\right) \\ $$$$\mathrm{into}\:\left({iv}\right)\:\Rightarrow\:\mathrm{5}{k}\:+\:\mathrm{2}\:\equiv\:\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{7}\right)\: \\ $$$$\:\Rightarrow\:\mathrm{5}{k}\:\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\:\right)\:\Leftrightarrow\:\:{k}\:\equiv\:\mathrm{5}\left(\:\mathrm{mod}\:\mathrm{7}\right)\:\Rightarrow\:{k}\:=\:\mathrm{7}{t}\:+\:\mathrm{5}\:,\:{t}\:\in\:\mathbb{N} \\ $$$$\mathrm{hence}\:{x}\:=\:\mathrm{5}\left(\mathrm{7}{t}\:+\:\mathrm{5}\right)\:+\:\mathrm{2}\:=\:\mathrm{35}{t}\:+\:\mathrm{27} \\ $$$${x}\:=\:\mathrm{35}{t}\:+\:\mathrm{27}\:, \\ $$$$\:\:\: \\ $$$$\: \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *