Question

A two digit number is such that product of its digits is $18$. When $63$ is subtracted from the no, the digits interchanges then places. Find the number.

Answer 1

Let the ten’s digit be $x$

Then, the unit's digit$=\frac{18}{x}$

Number$=10x+\frac{18}{x}$

And number obtained by inter changing the digits $=10\times \frac{18}{x}+x$

According to the question,

$10x+\frac{18}{x}-\frac{180}{x}-x=63$

$\Rightarrow 9x-\frac{162}{x}=63$

$\Rightarrow 9x^2-63x-162=0$

$\Rightarrow x^2-7x-18=0$

$\Rightarrow x^2-9x+2x-18=0$

$\Rightarrow x(x-9)+2(x-9)=0$

$\Rightarrow (x-9)(x+2)=0$

Ignoring $x=-2$

$\therefore x=9$

Hence the required number is $10\times 9+\frac{18}{9}=90+2=92$