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\(3^-1 mod 59\) how do negative exponents and modular arithmetic work?
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3^-1 is the multiplicative inverse of 3 in mod 59 so you are looking for an x value that makes this equation true 3x = 1 (mod 59)
ah, so 3^-5 would be 3^5x?
what does it mean when it says \(3^{-5}\cong 20\)
mod 59, but that isn't important (I don't think)
we know that \(3^{-1}\) is \(\dfrac{1}{3}\) in modulo \(59\) : \[\dfrac{1}{3} = \dfrac{20}{60}\equiv \dfrac{20}{1} = 20\]
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oh yeah, you showed me this trick a week ago
and so \(3^{-5}=(3^{-1})^5\)
which is congruent to 17, which is what my professor got. awesome. thank you so much ganesh. sorry about being a jerk jim
and the other 2
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