An algebra problem by Deepa Indu

as per Ques. we have
X*Y=12 and 10X+Y+36=10Y+X solving these we have X=2 and Y=6

$t=ten's~digit; u=unit's~digit$

$tu=12$ or $t=\dfrac{12}{u}$ $\implies$ $\boxed{1}$

$10t+u+36=10u+t$ $\implies$ $t-u+4=0$ $\implies$ $\boxed{2}$

Substitute $\boxed{1}$ in $\boxed{2}$ , we get

$u^2-4u-12=0$

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

$u=6$ or $u=-2$

Disregard the negative value. So, $u=6$ . It follows that $t=2$ . So the original number is $26$ .

When I saw that all the option when their digit
are multiplied it comes 12. Then I thought to add them 36 as the condition was given that by adding the no. with 36 its digits will interchange the digits. And finally I got the answer 26.