Freeflows
Modulo Congruence Puzzle 2
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} \pmod{m} $. She tries the example $ a=2 $, $ b=3 $, and $ m=7 $. Let $ L $ 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 $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$