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# Need Help Urgently

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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 $(ab)^{-1} \equiv 2$ (mod $n$). Given $a+b$ is invertible, what is the remainder when $(a+b)^{-1}(a^{-1}+b^{-1})$ is divided by $n$?

Jun 16, 2021

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Multiply $${(a+b)}^{-1}({a}^{-1}+{b}^{-1})$$ by ab

$${(a+b)}^{-1}({a}^{-1}+{b}^{-1})$$(ab) =$${(a+b)}^{-1}({a}^{-1}(ab)+{b}^{-1}(ab))$$

$$={(a+b)}^{-1}(b+a)$$

= 1 (mod n)

This shows that $${(a+b)}^{-1}({a}^{-1}+{b}^{-1})$$ = $${(ab)}^{-1}$$= 2 (mod n)

Jun 17, 2021