Question

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

Answer 1

Let $n(a,b)$ be two digit number.

$n(a,b)=10a+b$

Given, $\frac{n(a,b)}{ab}=\frac{8}{3}$

$\frac{10a+b}{ab}=\frac{8}{3}\Rightarrow 30a+3b=8ab$

Given that $n(a,b)-n(b,a)=18$

$9(a-b)=18$

$a-b=2$

$a=b+2$

$30a+3(a-2)=8a(a-2)$

$33a-6=8a^2-16a$

$8a^2-49a+6=0$

$(8a-1)(a-6)=0$

$\therefore a=6,a\ne \frac{1}{8}$

$b=a-2=4$

$64$ is the required number.