Question

A two digit number is such that the product of its digit is $30$, when $9$ is substracted from the number, the digits interchange their places. Find the number.

Answer 1

Let the digit in the ten's place be $x$.
Given that the product of two numbers is $30$.
$\therefore$ Unit digit$=\frac{30}{x}$
Number $=10x+\frac{30}{x}$
When the digit interchange their place, the new number
$=10\times \frac{30}{x}+x$
According to given condition, we have Old number $-9=$ New number
$=10\times \frac{30}{x}+x$

$10x+\frac{30}{x}-9=10\times \frac{30}{x}+x$

$10x^2+30-9x=300+x^2$

$9x^2-9x-270=0$

$x^2-x-30=0$

$(x+5)(x-6)=0$

$x=-5,x=6$

$x=-5$ is rejected as it is negative value.

$\therefore x=6$

Put this value of x in $\frac{30}{x}$ or $\frac{30}{6}=5$

$\therefore$ No. of $65$. Thus the required No. is $65$.