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I've been stuck on this one for a while, I just don't know what to do with all of this info :(

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 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.

 Dec 8, 2019
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Let us see if we can make some sense out of this !!

 

I don't think that ( a +b)^-1 is always congruent to a^-1 + b^-1 mod m.

 

Let us use your example:

 

(2 + 3)^-1 mod 7 =

3 mod 7 = 3, which is your L

 

While 2^-1  +  3^-1 mod 7 =

              4      +     5=9 mod 7 =2, which is your R

 

So, L - R = 3 - 2 =1

 Dec 8, 2019

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