Find three consecutive odd numbers whose sum is 147.

Let the first odd number be $\mathrm{x}$.

Let the second odd number be $(\mathrm{x}+2)$.

Let the third odd number be $(\mathrm{x}+4)$.

$\therefore x+(x+2)+(x+4)=147$

$\Rightarrow x+x+2+x+4=147$

$\Rightarrow 3 x+6=147$

$\Rightarrow 3 x=147-6$

$\Rightarrow 3 x=141$

$\Rightarrow x=\frac{141}{3}=47$

Therefore, the first odd number is $47 .$

$S$ econd odd number $=(\mathrm{x}+2)=(47+2)=49$

$T$ hird odd number $=(\mathrm{x}+4)=(47+4)=51$