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.)
I know that if $13^{-1} \equiv 29 \pmod{47}$, then $13 \cdot 29 \equiv 1 \pmod{47}$, by the definition of the inverse modulo 47. I also know that $34 \equiv -13 \pmod{47}$, since $34 + 13 = 47$. Therefore, $34^{-1} \equiv -13^{-1} \pmod{47}$, by the properties of inverses. But I also know that $-13^{-1} \equiv -29 \pmod{47}$, since multiplying both sides of the congruence by -1 preserves the congruence. So, $34^{-1} \equiv -29 \pmod{47}$. However, I want to find a residue modulo 47, which means a number between 0 and 46, inclusive. I can do this by adding 47 to -29, since that does not change the congruence. So, $34^{-1} \equiv -29 + 47 \equiv 18 \pmod{47}$. 