Find two-digit number the quotient of whose division by the product of its digits is equal to \[\dfrac{8}{3}\]
and the difference between the required number and the number consisting of the same digits written in the reverse
order is $18$

Hint: We will suppose unit digit and tens digit to get a two digit number (original number). Further we will apply the first condition to solve the problem .Thereafter, we will replace unit digit into tens digit and tens digit into unit digit to get the reverse number of the original number. Then we will apply the second condition to solve the problem. Later on we will find unit digit and tens digit then put in the original number to get the desired result.


Complete step by step solution:
We will suppose a two digit number where
Unit digit be $x$.
And tens digit be $y$
Then the original number will be $10y + x$.
When we interchange the digit, then number $ = 10x + y$
Now, will solve according to question
Case-1: The quotient of whose division by the product of its digits is equal to \[\dfrac{8}{3}\]
\[\dfrac{{10y + x}}{{xy}} = \dfrac{8}{3}\]
Cross multiplying the number, we will get
$3(10y + x) = 8xy$
$30y + 3x = 8xy\,\,\,\,\,\,\,\,\,\,\,\,\,....(1)$
Case 2: The difference between the required number and the number consisting of the same digits is $18$
$(10y + x) - (10x + y) = 18$
$10y + x - 10x - y = 18$
$9y - 9x = 18$
$9(y - x) = 18$
$y - x = \dfrac{{18}}{9}$
$y - x = 2$
$ \Rightarrow y - 2 = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(ii)$
Now, we will substitute the value of $x$in equation\[\left( i \right)\], we have
$30y + 3(y - 2) = 8y(y - 2)$
$30y + 3y - 6 = 8{y^2} - 16y$
$33y - 6 = 8{y^2} - 16y$
\[ \Rightarrow 8{y^2} - 16y - 33y + 6 = 0\]
$ \Rightarrow 8{y^2} - 49y + 6 = 0$
Splitting middle term in above equation
$ \Rightarrow 8{y^2} - 48y - y + 6 = 0$
$ \Rightarrow 8y(y - 6) - 1(y - 6) = 0$
$ \Rightarrow (8y - 1)(y - 6) = 0$
$ \Rightarrow 8y - 1 = 0$ $y - 6 = 0$
$8y = 0 + 1$ $y = 0 + 6$
$y = \dfrac{1}{8}$ $y = 6$
So, $y \ne \dfrac{1}{8}$ So, $y = 6$
Therefore, we will put the value of $y$in equation (ii)
$y - 2 = x$
$6 - 2 = x$
$4 = x$
The original number $ = 10y + x$
\[ = 10 \times 6 + 4\]
$ = 60 + 4$
$ = 64$
Hence, the required number is $64$


Note: The approach to these questions is made by repeatedly reading the statements of the question because the equation by which we will be able to solve these questions is hidden within that statement only.