Question

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

Answer 1

Hint: Approach the solution by applying the conditions to the assumed value.
In a two digit number, let us consider ten’s digit place as $x$ and unit’s digit place as $y$
Therefore two digit number will be = $10x+y$
Given product of the two digits = $12$
$\therefore$$xy=12$
$\Rightarrow y={\displaystyle \frac{12}{x}}\to (1)$
And also added $36$ to the two digit number
$\Rightarrow 10x+y+36$
Here after adding $36$ to the two digit number, the digits interchanges the places
$$\begin{gathered} \Rightarrow 10x+y+36=10y+x \\ \Rightarrow 9x-9y+36=0 \\ \Rightarrow x-y+4=0\to (2) \end{gathered}$$
From $(1)\mathrm{\&}(2)$ we get
$$\begin{gathered} \Rightarrow x+4={\displaystyle \frac{12}{x}} \\ \Rightarrow x^2+4x-12=0 \\ \Rightarrow (x+6)(x-2)=0 \end{gathered}$$
Here $x=-6$ and $x=2$ [Based on the given condition and the condition we have $x=-6$ value is rejected.]
Now on rejecting, x=-6, we have x value as $x=2$
$$\begin{gathered} \Rightarrow y={\displaystyle \frac{12}{x}} \\ \Rightarrow y={\displaystyle \frac{12}{2}} \\ \Rightarrow y=6 \end{gathered}$$
Here the required two digit number is $10x+y$
So, on substituting x, y values we get the number as
$\Rightarrow 10(2)+6=26$
Therefore the required two digit number is $26$.
NOTE: Here we should not ignore the condition of interchanging the places of digits. And here we have to consider the x value based on the conditions given (according to the solution).