Question

How many solutions does the equation 6x = 4 have in (b) ${\mathbf{\mathbb{Z}}}_{\mathbf{8}}$?

Short answer

The solutions of the equation are x = [2]₈ and x = [6]₈.

Step-by-step solution

Write the elements

The set of ${\mathrm{\mathbb{Z}}}_8$contains the elements $\left\{0,1,2,3,4,5,6,7\right\}$. As the number 8 is not prime, so the given equation can be or cannot be exits.

Find the solution

Directly substitute the value into the equation asfollows:

$$\begin{gathered} 6·{\left[0\right]}_8={\left[0\right]}_8 \\ 6·{\left[1\right]}_8={\left[8\right]}_8 \\ 6·{\left[2\right]}_8={\left[12\right]}_8 \\ ={\left[4\right]}_8 \end{gathered}$$

$$\begin{gathered} 6·{\left[3\right]}_8={\left[18\right]}_8 \\ ={\left[2\right]}_8 \\ 6·{\left[4\right]}_8={\left[24\right]}_8 \\ ={\left[0\right]}_8 \end{gathered}$$

It is observed that the number of unique solutions is obtained by modulus divided by multiple of 6 that gives [0]₈.This implies that 8/4 = 2. Therefore, there are two solutionstothe given equation.

The remaining solution is obtained by adding or subtractingthemultiple of [4]₈ from solution [2]₈.

This implies:

$$\begin{gathered} 6·{\left[6\right]}_8={\left[36\right]}_8 \\ ={\left[4\right]}_8 \end{gathered}$$

The solutions to the equation arex = [2]₈ and x = [6]₈.