Question

A two digit positive number is such that the product of its digits is $6$. If $9$ is added to the number, the digits interchange their places. Find the number.

Answer 1

Let the required two digit number be $10x+y$

Given: $xy=6$ and $10x+y+9=10y+x$

$10x-x+y-10y+9=0$

$\Rightarrow 9x-9y+9=0$

$\Rightarrow x-y+1=0$

$\Rightarrow y=x+1$

$\because xy=6$ (given)

$\Rightarrow x(x+1)=6$

$\Rightarrow x^2+x-6=0$

$\Rightarrow x^2+3x-2x-6=0$

$\Rightarrow (x+3)(x-2)=0$

Rejecting $x=-3$

When $x=2,y=x+1=2+1=3$

$\therefore$ The required two digit number $=10x+y$

$=10\times 2+3$

$=23$.