A two-digit number is such that the product of its digits is 8. When 18 is subtracted from the number, the digits interchange their places. Find the number.


Given:

The product of the digits of a two-digit number is 8.

When 18 is subtracted from 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,

$xy=8$-----(1)

$10x+y-18=10y+x$

$10x-x+y-10y=18$

$9x-9y=18$

$9(x-y)=18$

$x-y=2$

$x=2+y$

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

$(2+y)y=8$

$2y+y^2=8$

$y^2+2y-8=0$

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

$y^2+4y-2y-8=0$

$y(y+4)-2(y+4)=0$

$(y-2)(y+4)=0$

$y-2=0$ or $y+4=0$

$y=2$ or $y=-4$

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

$y=2$, then $x=2+2=4$

The required number is $10x+y=10(4)+2=40+2=42$.

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Updated on: 10-Oct-2022

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