A natural number, when increased by 12,

Let the natural number be x.

According to the question,

$x+12=\frac{160}{x}$

On multiplying by $x$ on both sides, we get

$\Rightarrow \quad x^{2}+12 x-160=0$

$x^{2}+(20 x-8 x)-160=0$

$\Rightarrow \quad x^{2}+20 x-8 x-160=0$ [by factorisation method]

$\Rightarrow \quad x(x+20)-8(x+20)=0$

$\Rightarrow \quad ; \quad(x+20)(x-8)=0$

Now, x + 20 = 0 ⇒ x = – 20 which is not possible because natural number is always greater than zero and x-8 = 0⇒x = 8.

Hence, the required natural number is 8.