Question

The sum of the squares of the digits constituting a two-digit positive number is $2.5$ times as large as the sum of its digits and is larger by unity than the trebled product of its digits. Find the number.

Answer 1

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

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

Given that, $a^2+b^2=\frac{5}{2}(a+b)\to \left(1\right)$

and $a^2+b^2=3ab+1\to \left(2\right)$

$\therefore 5(a+b)=2(3ab+1)$

$25(a+b{)}^2=4(3ab+1{)}^2$

$25(a^2+b^2+2ab)=46(9a^2{b}^2+6ab+1)$

$25(5ab+1)=36a^2{b}^2+24ab+4$

$36a^2{b}^2-101ab-21=0$

$\therefore ab=\frac{101\pm \sqrt{{101}^2+4\times 36\times 21}}{72}$

$=\frac{101+115}{72}=3$

$\therefore a^2+b^2=1$

$a-b=\sqrt{a^2+b^2-2ab}=\pm 2$

$a+b=\sqrt{a^2+b^2+2ab}=4$

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

(Not possible)

$\therefore 13$ is required number.