A two-digit number is such that the product of its digits is 8.

Question:

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.

Solution:

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