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.
Let the tens digit be $x$, then, the unit digits $=\frac{8}{x}$
Therefore, number $=\left(10 x+\frac{8}{x}\right)$
And number obtained by interchanging the digits $=\left(10 \times \frac{8}{x}+x\right)$
Then according to question
$\left(10 x+\frac{8}{x}\right)-\left(10 \times \frac{8}{x}+x\right)=18$
$\left(10 x+\frac{8}{x}\right)-\left(10 \times \frac{8}{x}+x\right)=18$
$\frac{\left(10 x^{2}+8\right)-\left(80+x^{2}\right)}{x}=18$
$\frac{10 x^{2}+8-80-x^{2}}{x}=18$
$\frac{9 x^{2}-72}{x}=18$
$9 x^{2}-72=18 x$
$9 x^{2}-18 x-72=0$
$9\left(x^{2}-2 x-8\right)=0$
$x^{2}-2 x-8=0$
$x^{2}-4 x+2 x-8=0$
$x(x-4)+2(x-4)=0$
$(x-4)(x+2)=0$
$(x-4)=0$
$x=4$
Or
$(x+2)=0$
$x=-2$
So the digit can never be negative.
Therefore,
When $x=4$ then the unit digits is
$=\frac{8}{x}$
$=\frac{8}{4}$
$=2$
And therefore the number is
$=\left(10 x+\frac{8}{x}\right)$
$=(10 \times 4+2)$
$=42$
Thus, the required number be 42
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