Question

Find the two-digit number, given: the quotient of the required number and the sum of its digits is $8$; the quotient of the product of its digits and that sum is $\frac{14}{9}$.

Answer 1

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

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

Given that, $\frac{n(a,b)}{a+b}=8$

$10a+b=8a+8b$

$2a=7b$

$\frac{\text{Product of digits}}{\text{Sum of digits}}=\frac{14}{9}$

$\frac{ab}{a+b}=\frac{14}{9}$

$9ab=14a+14b$

$18a^2=7(14a+4a)(\because 7b=2a)$

$a^2=7a$

$a=0$ (or) $7$ but $a$ can't be $0$

$a=7\Rightarrow b=2(\because b=\frac{2a}{7})$

$72$ is required number.