Question

Use Exercise 13 to solve the following equations.

(a) $15x=9$in ${\mathrm{\mathbb{Z}}}_{18}$(b) $25x=10$in${\mathrm{\mathbb{Z}}}_{65}$

Short answer

(a) The solution set is $\left\{\left[3\right],\left[9\right],\left[15\right]\right\}$.

(b)The solution set is$\left\{\left[3\right],\left[16\right],\left[29\right],\left[42\right],\left[55\right]\right\}$.

Step-by-step solution

Part (a)

The given equation in ${\mathrm{\mathbb{Z}}}_{18}$is $15x=9$. We have $\left(15,18\right)=3$,and 3 divides 9, which is the constant in the given equation. Then the solution set of the given equation can be found as follows:

$\begin{array}{rcl}15·\left(-1\right)+18·1& =& 3\\ 15& =& 3·5\\ 9& =& 3·3\\ 18& =& 3.6\end{array}$

From the obtained equations, the solution set will be $\left\{\left[3\right],\left[9\right],\left[15\right]\right\}$.

Part (b)

The given equation in ${\mathrm{\mathbb{Z}}}_{65}$ is $25x=10$.We have $\left(25,65\right)=5$,and 5 divides 10, which is the constant in the given equation. Then the solution set of the given equation can be found as follows:

$\begin{array}{rcl}25·8+65·\left(-3\right)& =& 5\\ 25& =& 5·5\\ 10& =& 5·2\\ 65& =& 5·13\end{array}$

From the obtained equations, the solution set will be $\left\{\left[3\right],\left[16\right],\left[29\right],\left[42\right],\left[55\right]\right\}$.