The solution set of 8x equiv 6 pmod{14},x in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the linear congruence $8x \equiv 6 \pmod{14}$.This is equivalent to $8x - 6 = 14k$ for some integer $k \in \mathbb{Z}$.Dividing by $2$,we ge

Vedclass · Accepted Answer

$[6] \cup [13]$

Vedclass · Answer

(C) Given the linear congruence $8x \equiv 6 \pmod{14}$.This is equivalent to $8x - 6 = 14k$ for some integer $k \in \mathbb{Z}$.Dividing by $2$,we get $4x - 3 = 7k$,which implies $4x \equiv 3 \pmod{7}$.Multiplying by the modular inverse of $4$ modulo $7$,which is $2$ (since $4 	imes 2 = 8 \equiv 1 \pmod{7}$):$2(4x) \equiv 2(3) \pmod{7} \implies 8x \equiv 6 \pmod{7} \implies x \equiv 6 \pmod{7}$.Thus,$x$ can be written as $x = 7n + 6$ for $n \in \mathbb{Z}$.For $n = 0, x = 6$. For $n = 1, x = 13$. For $n = 2, x = 20$,and so on.The solutions are $x \in \{ \dots, -1, 6, 13, 20, 27, \dots \}$.This set can be represented as the union of two equivalence classes modulo $14$: $[6] = \{ \dots, -8, 6, 20, 34, \dots \}$ and $[13] = \{ \dots, -1, 13, 27, 41, \dots \}$.Therefore,the solution set is $[6] \cup [13]$.

The solution set of $8x \equiv 6 \pmod{14}$,$x \in \mathbb{Z}$,is

Similar Questions

The number of roots of the equation $|x|^2 - 7|x| + 12 = 0$ is

Solve the equation $21x^{2} - 28x + 10 = 0$.

If $p, q, r$ are in $A.P.$ and are positive,the roots of the quadratic equation $px^2 + qx + r = 0$ are all real for

Solve the equation $-x^{2}+x-2=0$.

How many roots does the equation $x - \frac{2}{x - 1} = 1 - \frac{2}{x - 1}$ have?

Vedclass Test Series

Exam Paper Generator

Online Exam Module