Menu Close

Solve-the-system-of-congruences-x-2-mod-3-x-5-mod-7-




Question Number 80504 by Rio Michael last updated on 03/Feb/20
Solve the system of congruences  x ≡ 2 (mod 3)  x ≡ 5( mod 7)
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{congruences} \\ $$$${x}\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$$${x}\:\equiv\:\mathrm{5}\left(\:\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\: \\ $$
Commented by mr W last updated on 03/Feb/20
x=21n+5
$${x}=\mathrm{21}{n}+\mathrm{5} \\ $$
Commented by Rio Michael last updated on 03/Feb/20
please check this sir:   x ≡ 2(mod 3) ⇒ x = 3k + 2   3k + 2 ≡ 5(mod 7) ⇒ 3k ≡ 3(mod 7) k has no   solution here sir!  for chinese remainder theorem   R_1  = 7 and R_2  = 3   7X_1 ≡ 1 (mod 3) ⇒ X_1  ≡ 1 (mod 3)  3X_2  ≡ 1 (mod 7) ⇒ X_2  ≡ 5(mod 7)  x ≡ R_1 a_1 X_1  + R_2 a_2 X_2 (mod n_1 n_2 )  x ≡ (7)(2)(1) + (3)(5)(5) (mod 21)  x ≡ 5(mod 21)
$${please}\:{check}\:{this}\:{sir}: \\ $$$$\:{x}\:\equiv\:\mathrm{2}\left({mod}\:\mathrm{3}\right)\:\Rightarrow\:{x}\:=\:\mathrm{3}{k}\:+\:\mathrm{2}\: \\ $$$$\mathrm{3}{k}\:+\:\mathrm{2}\:\equiv\:\mathrm{5}\left({mod}\:\mathrm{7}\right)\:\Rightarrow\:\mathrm{3}{k}\:\equiv\:\mathrm{3}\left({mod}\:\mathrm{7}\right)\:{k}\:{has}\:{no}\: \\ $$$${solution}\:{here}\:{sir}! \\ $$$${for}\:{chinese}\:{remainder}\:{theorem} \\ $$$$\:{R}_{\mathrm{1}} \:=\:\mathrm{7}\:{and}\:{R}_{\mathrm{2}} \:=\:\mathrm{3} \\ $$$$\:\mathrm{7}{X}_{\mathrm{1}} \equiv\:\mathrm{1}\:\left({mod}\:\mathrm{3}\right)\:\Rightarrow\:{X}_{\mathrm{1}} \:\equiv\:\mathrm{1}\:\left({mod}\:\mathrm{3}\right) \\ $$$$\mathrm{3}{X}_{\mathrm{2}} \:\equiv\:\mathrm{1}\:\left({mod}\:\mathrm{7}\right)\:\Rightarrow\:{X}_{\mathrm{2}} \:\equiv\:\mathrm{5}\left({mod}\:\mathrm{7}\right) \\ $$$${x}\:\equiv\:{R}_{\mathrm{1}} {a}_{\mathrm{1}} {X}_{\mathrm{1}} \:+\:{R}_{\mathrm{2}} {a}_{\mathrm{2}} {X}_{\mathrm{2}} \left({mod}\:{n}_{\mathrm{1}} {n}_{\mathrm{2}} \right) \\ $$$${x}\:\equiv\:\left(\mathrm{7}\right)\left(\mathrm{2}\right)\left(\mathrm{1}\right)\:+\:\left(\mathrm{3}\right)\left(\mathrm{5}\right)\left(\mathrm{5}\right)\:\left({mod}\:\mathrm{21}\right) \\ $$$${x}\:\equiv\:\mathrm{5}\left({mod}\:\mathrm{21}\right) \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published.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