# 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

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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).

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).

Last updated date: 29th Sep 2023

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