Question

The solution sets of $8x=6\left(\mathrm{mod}14\right),x\in \mathrm{\mathbb{Z}}$, are

• A. $\left[8\right]\cup \left[6\right]$
• B. $\left[8\right]\cup \left[14\right]$
• C. $\left[6\right]\cup \left[13\right]$
• D. $\left[8\right]\cup \left[6\right]\cup \left[13\right]$

Answer 1

The correct option is C

$\left[6\right]\cup \left[13\right]$

Solve for the given set:

The given equation: $8x=6\left(\mathrm{mod}14\right),x\in \mathrm{\mathbb{Z}}$.

$$\begin{gathered} \Rightarrow 8x-6=14k\left[k\in \mathrm{\mathbb{Z}}\right] \\ \Rightarrow 8x=14k+6 \\ \Rightarrow x=\frac{1}{4}\left(7k+3\right) \end{gathered}$$

For $k=3$.

$x=\frac{1}{4}\left[7\left(3\right)+3\right]=\frac{24}{4}=6$.

For $k=7$.

$x=\frac{1}{4}\left[7\left(7\right)+3\right]=\frac{52}{4}=13$.

For $k=11$.

$x=\frac{1}{4}\left[7\left(11\right)+3\right]=\frac{80}{4}=20$.

For $k=15$.

$x=\frac{1}{4}\left[7\left(15\right)+3\right]=\frac{108}{4}=27$.

Thus, the solution set can be given by, $x=\left\{6,13,20,27.....\right\}$.

$$\begin{gathered} \Rightarrow x=\left\{6,20,34,48,....\right\}\cup \left\{13,27,41,55,....\right\} \\ \Rightarrow x=\left[6\right]\cup \left[13\right] \end{gathered}$$

Therefore, the solution sets of $8x=6\left(\mathrm{mod}14\right),x\in \mathrm{\mathbb{Z}}$, are $\left[6\right]\cup \left[13\right]$.

Hence, the option (C) is correct.