The solution set of the congruence 8x equiv 6 | vedclass

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

Question

(A) The congruence $8x \equiv 6 \pmod{14}$ can be written as $8x - 6 = 14k$ for some integer $k \in \mathbb{Z}$.Dividing by $2$,we get $4x - 3 = 7k$,w

Vedclass · Accepted Answer

$[6] \cup [13]$

Vedclass · Answer

(A) The congruence $8x \equiv 6 \pmod{14}$ can be written as $8x - 6 = 14k$ for some integer $k \in \mathbb{Z}$.Dividing by $2$,we get $4x - 3 = 7k$,which implies $4x \equiv 3 \pmod{7}$.To solve $4x \equiv 3 \pmod{7}$,we multiply by the modular inverse of $4$ modulo $7$. Since $4 	imes 2 = 8 \equiv 1 \pmod{7}$,the inverse is $2$.Multiplying both sides by $2$: $8x \equiv 6 \pmod{7}$,which simplifies to $x \equiv 6 \pmod{7}$.This means $x$ can be written in the form $x = 7n + 6$ for $n \in \mathbb{Z}$.For $n = 0, x = 6$. For $n = 1, x = 13$. For $n = 2, x = 20$,and so on.The solution set consists of integers congruent to $6 \pmod{7}$ or $13 \pmod{7}$,which can be represented as the union of equivalence classes $[6]$ and $[13]$ modulo $14$.

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

Similar Questions

If $a, b, c$ are distinct real numbers and $a^3 + b^3 + c^3 = 3abc$,then the equation $ax^2 + bx + c = 0$ has two roots,out of which one root is

If the roots of the equation $(a^2 + b^2)t^2 - 2(ac + bd)t + (c^2 + d^2) = 0$ are equal,then:

Let $x_1, x_2, x_3 \in \mathbb{R} \setminus \{0\}$,$x_1 + x_2 + x_3 \neq 0$ and $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$. Then $\frac{1}{x_1^n + x_2^n + x_3^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n}$ holds good for:

If $a > 0$ and $b > 0$,what is the nature of the roots of the equation $a - bx - x^2 = 0$?

If $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots \infty}}}$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module