A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Let the unit's digit be $=x$
Since product of the digits $= 18$
$\implies$ $\text{Unit's digit} \times \text{Ten's digit} = 18$
$\implies$ $\text{Ten's digit}= \dfrac{18}{\text{Unit's digit}} = \dfrac{18}{x}$
$\therefore$ The two digit number is $= \dfrac{180}{x}+x$
On subtracting 63 from this number, $\text{Unit's digit} = \dfrac{18}{x}$ and $\text{Ten's digit}=x$
$\therefore$ New two digit number is $= 10x + \dfrac{18}{x}$
$\therefore$ As per question,
$\dfrac{180}{x}+x-63=10x+\dfrac{18}{x}$
i.e. $180+x^2-63x=10x^2+18$
i.e. $9x^2+63x-162=0$
i.e. $x^2+7x-18=0$
i.e. $x^2+9x-2x-18=0$
i.e. $x\left(x+9\right) - 2 \left(x+9\right)=0$
i.e. $\left(x+9\right) \left(x-2\right)=0$
i.e. $x=-9$ or $x=2$
Since unit's digit cannot be negative, $x=-9$ is not a valid answer.
$\therefore$ Unit's digit $=2$
$\therefore$ Ten's digit $= \dfrac{18}{2}=9$
$\therefore$ The number is $9\times10+2 = 92$