A natural number, when increased by 12,

It is given that a number when increased by 12 becomes 160 times its reciprocal.

We have to find the number.

Let the number be x

Reciprocal of $x=\frac{1}{x}$

According to the question

$x+12=160 \times \frac{1}{x}$

$x^{2}+12 x=160$

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

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

$x(x+20)-8(x+20)=0$

$(x+20)(x-8)=0$

$x=-20,8$

since $-20+8 \neq \frac{160}{-20}$

Therefore $x=8$