The solution of $7x≡3(mod\, 5) $ is : |
$x≡5(mod\, 5)$ $x≡4(mod\, 5)$ $x≡1(mod\, 5)$ $x≡2(mod\, 5)$ |
$x≡4(mod\, 5)$ |
The correct answer is Option (2) → $x≡4(mod\, 5)$ Given, $⇒7x≡3(mod\, 5)$ $⇒2x≡3(mod\, 5)$ $⇒2x×3≡3×3(mod\, 5)$ $⇒6x≡9(mod\, 5)$ $⇒x≡4(mod\, 5)$ |