Question

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

Answer 1

Let $x$ be the digit at unit place.

It is given that the product of the digits is $12$, therefore, the digit $y$ at tens place is:

$xy=12\Rightarrow y=\frac{12}{x}.......\left(1\right)$

Therefore,

Original number$=10\times \frac{12}{x}+x=\frac{120}{x}+x$

Reverse number$=10\times x+\frac{12}{x}=10x+\frac{12}{x}$

It is also given that when $36$ is added to this number the digits interchange their places, therefore,

$10x+\frac{12}{x}=\frac{120}{x}+x+36\Rightarrow 10x-x=\frac{120}{x}-\frac{12}{x}+36\Rightarrow 9x-\frac{108}{x}-36=0\Rightarrow 9x^2-36x-108=0\Rightarrow 9(x^2-4x-12)=0\Rightarrow x^2-4x-12=0\Rightarrow x^2-6x+2x-12=0\Rightarrow x(x-6)+2(x-6)=0\Rightarrow (x+2)=0,(x-6)=0\Rightarrow x=-2,x=6$

We reject the negative value of $x$ and take $x=6$, thus, digit at the unit place is $6$.

Now, using equation 1, digit at the tens place is $y=\frac{12}{6}=2$.

Since the unit place of the digit is $6$ and the tens place is $2$

Hence, the required number is $26$.