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.

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$
$\Rightarrow y = \dfrac{{12}}{x} \to (1)$
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
$\ \Rightarrow 10x + y + 36 = 10y + x \\ \Rightarrow 9x - 9y + 36 = 0 \\ \Rightarrow x - y + 4 = 0 \to (2) \\ \$
From $(1)\& (2)$ we get
$\ \Rightarrow x + 4 = \dfrac{{12}}{x} \\ \Rightarrow {x^2} + 4x - 12 = 0 \\ \Rightarrow (x + 6)(x - 2) = 0 \\ \$
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 $x = 2$
$\ \Rightarrow y = \dfrac{{12}}{x} \\ \Rightarrow y = \dfrac{{12}}{2} \\ \Rightarrow y = 6 \\ \$
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).