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.)
Looking at this problem, the first thing that I notice is that 13 + 37 = 47.
34 = -13 (mod 47)
13^(-1) = 29 (mod 47)
-13^(-1) = -29 (mod 47)
-29 = 18 (mod 47)
=^._.^=
13 + 37 = 47 ???
Oops, I meant 34 not 37.
13+34 = 47.
Thank you for noticing.
I figured that is what you meant.
Thx Guys!