Question

Find a two-digit number which exceeds by $12$ the sum of the squares of its digits and by $16$ the doubled product of its digits.

Answer 1

let the number be $10a+b$

From the given conditions

$10a+b=a^2+b^2+\mathrm{12......}\left(1\right)$

$10a+b=2ab+\mathrm{16..........}\left(2\right)$

$\left(1\right)-\left(2\right)⟹(a-b{)}^2=4⟹a-b=2$

$⟹10a+(a-2)=2\left(a\right)(a-2)+16$

$⟹2a^2-4a+16-10a-a+2=0$

$⟹a=6$

$⟹b=4$

So the number is $64$