Question

The sum of the squares of the digits of a two-digit number is $13$. If we subtract $9$ from that number, we get a number consisting of the same digits written in the reverse order. Find the number.

Answer 1

Let the digits be $x$ and $y$

The sum of square of the number is $13$

$\Rightarrow x^2+y^2=13$

Given that if $9$ is subtracted from the number, it get reversed

$\Rightarrow (10x+y)-9=10y+x$

$9x-9y=9$

$x-y=1$

Now, $(x-y{)}^2=x^2+y^2-2xy$
$\Rightarrow 1=13-2xy$
$\Rightarrow 2xy=12$
$\Rightarrow xy=6\Rightarrow y=\frac{6}{x}$
$\Rightarrow x-y=1$
$\Rightarrow x-\frac{6}{x}=1$
$\Rightarrow x^2-6=x$
$\Rightarrow x^2-x-6=0$
$\Rightarrow x^2+2x-3x-6=0$

$\Rightarrow x(x+2)-3(x+2)=0$

$\Rightarrow (x+2)(x-3)=0$

$\Rightarrow (x+2)=0$ and $(x-3)=0$

$\Rightarrow x=2,3$

Hence, the number is $32$.