Question

A number consists of two digits whose product is $18$. When $27$ is subtracted from the number, the digits change their places. Find the number.

Answer 1

Let the unit digit be $x$ so, tens digit will be $\frac{18}{x}.$

According to question,

$(x+\frac{18}{x}\times 10)-27=10\times x+\frac{18}{x}$

$\Rightarrow (x+\frac{180}{x})-27=10x+\frac{18}{x}$

Taking $x$ as LCM,

$\begin{array}{cc}{x}^2+180-27x& =10x^2+18\\ 9x^2-27x-162& =0\\ 9(x^2-3x-18)& =0\\ x^2-3x-18& =0\end{array}$

Now, we will factorize the above quadratic equation to find the roots,

$\begin{array}{cc}{x}^2-6x+3x-18& =0\\ x(x-6)+3(x-6)& =0\\ x+3=0\text{or}x-6& =0\\ x=-3\text{or}x& =6\end{array}$

Now if $x$ is negative then product will not be $+18$. Therefore $x=6$ is the only acceptable value of $x$.

Hence the number will be,

$\begin{array}{cc}x+\frac{180}{x}& =6+\frac{180}{6}\\ & =6+30\\ & =36\end{array}$