The sum of the squares of the digits constituting a two-digit number is $10$ and the product of the required number by the number consisting of the same digits written in the reverse order is $403$. Find the number.

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

Sum of squares of digits $=10$

$a^2+b^2=10$

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

$a>0,b>0$

$\left(number\right)\times \text{(number with same digits in reverse order)}=403$

$n(a,b)\times n(b,a)=403$

$(10a+b)(10b+a)=403$

$10(a^2+b^2)+\left(101ab\right)=403$

$100+101ab=403(\because a^2+b^2=10)$

$ab=3$

$(a+b{)}^2=a^2+b^2+2ab=16$

$a+b=4(\because a>0,b>0)$

$(a-b{)}^2=a^2+b^2-2ab=4$

$a-b=\pm 2$

$a=3,b=1$ (or) $a=1,b=3$

$\therefore$ $13$ (or) $31$ is the required number