Question

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

Answer 1

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

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

Given that $a^2+b^2=13$

$n(a,b)-9=n(b,a)$

$10a+b-9=10b+a$

$9(a-b)=9$

$\therefore a-b=1$

$a^2+b^2-2ab=1$

$2ab=12$ $(\because a^2+b^2=13)$

$ab=6$

$(a+b{)}^2=13+2\times 6$

$=25\Rightarrow a+b=5$ $(\because a\ge 1,b\ge 0)$

$a-b=1$ and $a+b=5$

$\therefore a=3,b=2$

$\therefore n(a,b)=32$ is required number.