Question

A two digit number is such that the product of the digits is $12$. When $36$ is added to the number, the digits interchange their places. The number is ------.

Answer 1

$\Rightarrow$ Let the unit digit in the number be $x$, then product of digits is $12$, the ten's digit in the number is $\frac{12}{x}$.

$\therefore$ The value of the number $=10\times \frac{12}{x}+x=\frac{120}{x}+x$

$\Rightarrow$ On reversing the digits, unit's digit becomes ten's digit and ten's digit becomes unit's digit and its value will become $10x+\frac{12}{x}$

Accroding to the question,

$\Rightarrow$ $10x+\frac{12}{x}=\frac{120}{x}+x+36$

$\Rightarrow$ $9x=\frac{120-12}{x}+36$

$\Rightarrow$ $9x=\frac{108}{x}+36$

$\Rightarrow$ $9x^2=108+36x$

$\Rightarrow$ $9x^2-36x-108=0$

$\Rightarrow$ $x^2-4x-12=0$

$\Rightarrow$ $x^2-6x+2x-12=0$

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

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

$\Rightarrow$ $x=6$ and $x=-2$

We can not have negative number in unit place

$\therefore$ $6$ is unit's place and in tens place we have $\frac{12}{6}=2$

$\therefore$ The required number $=26$