Question

A two digit number is such that the product of the digits is $14$. When $45$ is added to the number, then the digits are reversed. Find the number.

Answer 1

Let the tens digit be $x$. Then, units digit$=\frac{14}{x}$
With the above ten's and unit's digit, the number $=10x+\frac{14}{x}$

and number formed by reversing the digits $=10\times \frac{14}{x}+x$
According to the statement:

$10x+\frac{14}{x}+45=10\times \frac{14}{x}+x$

$10x+\frac{14}{x}+45=\frac{140}{x}+x$

$9x-\frac{126}{x}+45=0$

$9x^2+45x-126=0$

$x^2-5x-14=0$

$(x+7)(x-2)=0$

$\Rightarrow x=-7$ or $x=2$

$x$ can not be a negative

$\Rightarrow x=2$

Hence the required number $=(10\times 2)+\frac{14}{2}=27$