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Let n be a positive integer greater than or equal to 3 . Let a,b be integers such that  ab is invertible modulo n  and \((ab)^{-1}\equiv 2\pmod n\). Given a+b is invertible, what is the remainder when \((a+b)^{-1}(a^{-1}+b^{-1})\) is divided by n?

 Apr 15, 2021
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\((a+b)^{-1}(a^{-1}+b^{-1})=\frac{1}{ab} \)

 

the question has already told you that the remainder will be 2 when it is divided by n.

That is what mod means 

 

  mod means remainder.

 Apr 18, 2021

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