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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