Let $ S = w + 2w^2 + 3w^3 + \dots + 9w^9 $. Then \[wS = w^2 + 2w^3 + 3w^4 + \dots + 9w^{10}.\]Subtracting these equations, we get \[(1 - w) S = w + w^2 + w^3 + \dots + w^9 - 9w^{10}.\]Note that $ w^9 = \cos 360^\circ + i \sin 360^\circ = 1, $ so $ w^9 - 1 = 0 $. This factors as \[(w - 1)(w^8 + w^7 + \dots + w + 1) = 0.\]Since $ w \neq 1, $ \[w^8 + w^7 + \dots + w + 1 = 0.\]Hence, \begin{align*} (1 - w) S &= w + w^2 + w^3 + \dots + w^9 - 9w^{10} \\ &= w(1 + w + w^2 + \dots + w^8) - 9w \\ &= -9w, \end{align*}so \[S = -\frac{9w}{1 - w}.\]Now, \begin{align*} \frac{1}{1 - w} &= \frac{1}{1 - \cos 40^\circ - i \sin 40^\circ} \\ &= \frac{1 - \cos 40^\circ + i \sin 40^\circ}{(1 - \cos 40^\circ)^2 + \sin^2 40^\circ} \\ &= \frac{1 - \cos 40^\circ + i \sin 40^\circ}{2 - 2 \cos 40^\circ} \\ &= \frac{2 \sin^2 20^\circ + 2i \sin 20^\circ \cos 20^\circ}{4 \sin^2 20^\circ} \\ &= \frac{\sin 20^\circ + i \cos 20^\circ}{2 \sin 20^\circ} \\ &= \frac{\cos 70^\circ + i \sin 70^\circ}{2 \sin 20^\circ} \\ \end{align*}Then \begin{align*} \frac{1}{|S|} &= \left| \frac{1 - w}{9w} \right| \\ &= \frac{2 \sin 20^\circ}{9 |\cos 70^\circ + i \sin 70^\circ|} \\ &= \frac{2}{9} \sin 20^\circ.\end{align*} The final answer is $ 2 + 9 + 20 = \boxed{31} $.