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 $L-R$

 Aug 24, 2023

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.

 Aug 24, 2023

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.



--. .- 

GingerAle  Aug 25, 2023
edited by Guest  Aug 25, 2023

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. 




--. .- 

GingerAle  Aug 25, 2023
edited by GingerAle  Aug 25, 2023

Oh dear, here GA goes again…

It’s so obvious what is going on.



  • 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

0 Online Users