+0  
 
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4
avatar+145 

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 \(n\) ?

 Feb 14, 2021
edited by Jmaster10  Feb 14, 2021
 #1
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-2

The remainder is 4.

 Feb 14, 2021
 #2
avatar+145 
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It is not 4

 Feb 15, 2021
 #3
avatar+112533 
+1

 

I have never studied modulo mathematics except by looking at the odd question on this forum.

My setting out will be very poor but my logic should be ok.

 

 

\((ab)(ab)^{-1}\equiv 1\mod n\\ so\\ 2ab\equiv 1\;\;\mod n\\ 2\equiv \frac{1}{ab}\;\;\mod n\\~\\ \frac{1}{a+b}(\frac{1}{a}+\frac{1}{b})\mod n\\~\\ \equiv \frac{1}{a+b}(\frac{a+b}{ab})\mod n\\~\\ \equiv \frac{a+b}{ab(a+b)}\mod n\\~\\ \equiv \frac{1}{ab} \mod n\\~\\ \equiv 2 \mod n\\~\\ \)

 Feb 15, 2021
 #4
avatar+112533 
0

Now, I expect a polite response from you Jmaster10.

 

Thank you would be nice but more than just thank you would be better.  

I mean if you want to correct something or ask something, that would be good.

 Feb 15, 2021

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