Question

A two-digit number is such that the product of its digits is $20$. If $9$ is added to the number, the digits interchange their places. Find the number.

Answer 1

Let the $x$ be the number at units place and $y$ be the number at digits place no: will be $=x+10y$

$\therefore$ Reverse no: will be $10x+y$

Product of digit is $20$

$xy=20⟶\left(i\right)$

Now , original number $+9=$reverse no.

$x+10y+9=10x+y9x-9y=9x-y=1⟶\left(ii\right)$

Using formula $(x+y{)}^2=(x-y{)}^2+4xy$

${(x+y)}^2=1+4\left(20\right)x+y=\sqrt{81}x+y=9⟶\left(iii\right)$

Solve $\left(ii\right)$ & $\left(iii\right)$ ,

$\underset{–}{\begin{array}{c}x-y=1\\ x+y=9\end{array}}$

$2x=10$ $\Rightarrow x=5$

Putting value of $x$ in $\left(ii\right)$ ,

$5-y=1$

$y=4$

$\therefore$ no: is $=x+10y$

$=5+40$

$=45.$