Question

What two-digit number is less than the sum of the squares of its digits by $11$ and exceed their doubled product by $5$?

Answer 1

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

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

Given that, $n(a,b)+11=a^2+b^2$

$10a+b+11=a^2+b^2\to \left(1\right)$

$n(a,b)=2ab+5$

$10a+b=2ab+5\to \left(2\right)$

$\left(1\right)-\left(2\right)\Rightarrow 11=a^2+b^2-2ab-5$

$(a-b{)}^2=16$

$a-b=\pm 4\Rightarrow a=b+4\left(or\right)a=b-4$

$a=b+4$,

$10b+40+b=2b(b+4)+5$

$2b^2-3b-35=0$

$(2b+7)(b-5)=0$

$\therefore$ $b=5,a=9$

$a=b-4$

$10b-40+b=2b^2-8b+5$

$2b^2-19b+45=0$

$(2b-9)(b-5)=0$

$\therefore$ $b=5,a=1$

$15$ and $95$ are the required numbers.