A two-digit number is such that the product of digit is 12.

Let the tens digit be $x$ then, the unit digits $=\frac{12}{x}$

Therefore, number $=\left(10 x+\frac{12}{x}\right)$

And number obtained by interchanging the digits $=\left(10 \times \frac{12}{x}+x\right)$

Then according to question

$\left(10 x+\frac{12}{x}\right)+36=\left(10 \times \frac{12}{x}+x\right)$

$\left(10 x+\frac{12}{x}\right)+36=\left(\frac{120}{x}+x\right)$

$\left(10 x+\frac{12}{x}\right)-\left(\frac{120}{x}+x\right)+36=0$

$\frac{\left(10 x^{2}+12\right)-\left(120+x^{2}\right)+36 x}{x}=0$

$10 x^{2}+12-120-x^{2}+36 x=0$

$9\left(x^{2}+4 x-12\right)=0$

$x^{2}+4 x-12=0$

$x^{2}-2 x+6 x-12=0$

$x(x-2)+6(x-2)=0$

$(x-2)(x+6)=0$

$(x-2)=0$

$x=2$

Or

$(x+6)=0$

$x=-6$

So, the digit can never be negative.

Therefore,

When $x=2$ then the unit digits

$=\frac{12}{x}$

$=\frac{12}{2} .$

$=6$

And number

$=\left(10 x+\frac{12}{x}\right)$

$=(10 \times 2+6)$

$=26$

Thus, the required number be 26