$x, y$ are natural numbers such that $x > y$ . Also $x + y + x y = 80$ , then value of $x$ is?

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Solution

$x + y + x y = 80$
Let $x + y = n$
& $x y = 80 - n$
$\Rightarrow x (n - x) = 80 - n$
$\Rightarrow x^{2} - n x + (80 - n) = 0$
$D = n^{2} - 4 (80 - n) \geq 0$
$\Rightarrow n^{2} + 4 n - 320 \geq 0$
$\Rightarrow (n - 16) (n + 20) \geq 0$
$n \geq 16$
So, $x + y \geq 16$
$x y \geq 64$
Hit & Trail
$x = 26 y = 2$
(Satisfy condtn)

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