+0

-1
138
4
+161

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
-2

The remainder is 4.

Feb 14, 2021
#2
+161
0

It is not 4

Feb 15, 2021
#3
+113739
+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
+113739
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