Friday 24 May 2013

algebra precalculus - Finding Inverse in modulus m



I've been learning the Euclidean algorithm and came across this simple problem.



$101^{-1} (mod 203)$



So I attempted it as such:




$203 = 101(2) + 1$



So we got a gcd of 1, we can stop and do:



$1 = 203 - 101(2)$



And since it's mod 203, we have 101(2)



So shouldn't the answer be $2$?
My textbook says it's $201$,

help would be much appreciated, as this is confusing as ever.



Thanks.


Answer



Observe that



$$2\cdot101=202=-1\pmod{203}\implies (101)^{-1}=-2=201\pmod{203}$$



Or using the EA:




$$203=2\cdot101+1\implies1\cdot203+\color{red}{(-2)}\cdot101=1\implies $$



$$\implies101^{-1}=-2\pmod{203}=201\pmod{203}$$


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