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. Determine the number.

Given:

The product of the digits of a two-digit number is 12.

When 36 is added to the number the digits interchange their places.

 To do:

We have to find the number.

Solution:

Let the two-digit number be $10x+y$.

According to the question,

$xy=12$-----(1)

$10x+y+36=10y+x$

$10x-x+y-10y+36=0$

$9x-9y+36=0$

$9(x-y+4)=0$

$x-y+4=0$

$x=y-4$

Substituting the value of $x$ in equation (1), we get,

$(y-4)y=12$

$y^2-4y=12$

$y^2-4y-12=0$

Solving for $y$ by factorization method, we get,

$y^2-6y+2y-12=0$

$y(y-6)+2(y-6)=0$

$(y-6)(y+2)=0$

$y-6=0$ or $y+2=0$

$y=6$ or $y=-2$

Considering the positive value of $y$, we get,

$y=6$, then $x=y-4=6-4=2$

The required number is $10x+y=10(2)+6=20+6=26$. 

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Updated on: 10-Oct-2022

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