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≡2(modn). Given a+b is invertible, what is the remainder when (a+b)−1(a−1+b−1) is divided by n?
+118703 
This is my take:
(ab)−1=2modnWhat does this even mean, wellthe inverse of 7 is 1/7 because 7∗(1/7)=1so(ab)−1∗(ab)=1modn2ab=1modn (a+b)−1(a−1+b−1)=1a+b∗(b+aab)=1ab≡2modn
Coding:
(ab)^{-1}=2\mod n\\
\text{What does this even mean, well}\\
\text{the inverse of 7 is 1/7 because } 7*(1/7)=1\\
so\\
(ab)^{-1}*(ab)=1\mod n\\
2ab=1\mod n\\~\\
(a+b)^{-1}(a^{-1}+b^{-1})\\
=\frac{1}{a+b}*(\frac{b+a}{ab})\\
=\frac{1}{ab}\\
\equiv 2\mod n
