Question

The product of the digits of a two digit number is $36$. If $45$ is subtracted from the number, its digits interchange their places. Find the number.

Answer 1

$Let$ that no. be $10x+y$

According to question ,

$\left(x\right)\left(y\right)=36$ ..............1

And ,

$10x+y-45=10y+x$

or,$9x-9y=45$

or$x-y=5$ ...............2

put value of from eq. 2 to eq.1

So,$(y+5)\left(y\right)=36$

or ,$y^2+5y-36=0$

$y^2+9y-4y-36=0$

$y(y+9)-4(y+9)=0$

$(y+9)(y-4)=0$

so, $y=-9,4$

$y$ cannot be negative . so $y=4$

from eq.2

$x-4=5$

or $x=9$

So, that no. is $10\left(9\right)+4=94$