Given that $13^{-1} \equiv 29 \pmod{47}$, find $34^{-1} \pmod{47}$, as a residue modulo 47.  (Give a number between 0 and 46, inclusive.)
Note that $34 \equiv -13 \pmod{47}$.  Hence, \begin{align*}
34^{-1} &\equiv (-13)^{-1} \\
&\equiv (-1)^{-1} \cdot 13^{-1} \\
&\equiv (-1) \cdot 29 \\
&\equiv \boxed{18} \pmod{47}.
\end{align*}