Given $m\geq 2$ denote by $b^{-1}$ the inverse of $b\pmod{m}$. That is, $b^{-1}$ is the residue for which $bb^{-1}\equiv 1\pmod{m}$. Sadie wonders if $(a+b)^{-1}$ is always congruent to $a^{-1}+b^{-1}$ (modulo ). She tries the example $a=2$, $b=3$ and $m=7$. Let be the residue of $(2+3)^{-1}\pmod{7}$, and let $R$ be the residue of $2^{-1}+3^{-1}\pmod{7}$, where $L$ and $R$ are integers from $0$ to $6$ (inclusive). Find $L-R$

VanillaNavel Aug 24, 2023

#1**0 **

We have that 2−1≡4(mod7) and 3−1≡5(mod7), so L≡4+5≡1(mod7). Also, 2+3≡5(mod7), so (2+3)−1≡7−5≡2(mod7), so R≡2+5≡7(mod7). Therefore, L−R=6.

Here is a more detailed explanation of the steps involved:

We find that 2−1≡4(mod7) and 3−1≡5(mod7) using the Euclidean algorithm.

We find that L≡4+5≡1(mod7) by applying the congruence ab≡1(modm) to a=2, b=4, and m=7.

We find that R≡2+5≡7(mod7) by applying the congruence ab≡1(modm) to a=5, b=2, and m=7.

We find that L−R=6 by subtracting R from L.

Guest Aug 24, 2023

#5**-2 **

**For the record: The GingerRoot account is NOT under the control of GingerAle. **

**------------**

*Mr. !!BB! You finally got something right for once! Your imbecillic bullshit notions have finally produced something coherent and correct*.

**The answer is random bullshit; it has nothing to do with the question.**

**In any case, post #1 isn’t by any of the BBs. **

Mr. BB often answers questions correctly, but his solution methods are nebulous. The BB’s solve problems by intuitive reasoning. The reasoning is un-teachable and useless, because it only applies to the question at hand and not as a general solution method.

GA

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GingerAle
Aug 25, 2023

edited by
Guest
Aug 25, 2023

#7**-2 **

**For the record: The GingerBeer account is NOT under the control of GingerAle. **

**------------------------**

*Typical GA BS. GA uses guest or fake accounts to vent anger then goes like "For the record its not meeee" like we all know its you LMAO!*

*GingerBeer 2 hours ago*

*edited by Guest 2 hours ago*

**Oppsies .. we made a BIG booboo! Didn’t we... Yes we did!**

This was supposed to be a **guest post** and it was snagged by **GingerBeer**.

OH Darn it! Darn it to the hot place in the bowls of the Earth! How will I ever explain this?

I can’t, so I’ll just delete it. Too bad that it was in plain view for over two (2) hours.

GA

--. .-

GingerAle
Aug 25, 2023

#8**+2 **

Oh dear, here GA goes again…

It’s so obvious what is going on.

Timeline:

- GA posts insults on a guest account, probably planning to disavow them later.
- GA creates another “fake” account GingerBeer, which snagged the inconvenient post.
- GA logs off for a circus performance, plotting to cause more chaos later
- GA logs back on, sees the inconvenient post, and panics; so edits it into a “test post”
- For good measure, GA posts on her main account blaming her fictional nemesis for the BS she created

Guest Aug 25, 2023

edited by
Guest
Aug 25, 2023