Question

A two-digit positive number is such that the product its digits is $8.$ If $18$ is added to the number then the digits are reversed, then the original number is

Answer 1

Let the two digit number be $pq$
Now, given conditions
$p\times q=8\Rightarrow q=\frac{8}{p}\cdots \left(1\right)$
And
$(10p+q)+18=10q+p$
$\Rightarrow q-p=2$
Using equation $\left(1\right)$, we get
$\Rightarrow \frac{8}{p}-p=2\Rightarrow 8-p^2=2p\Rightarrow p^2+2p-8=0\Rightarrow p^2+4p-2p-8=0\Rightarrow (p-2)(p+4)=0\Rightarrow p=2,-4$
As the number is positive, so
$p=2\Rightarrow q=4$

Hence, the original number is $24$