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 $L-R$.

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LammettHash LammettHash · Answer 1 · 2020-07-24T17:28:08+02:00

[tex](2+3)^{-1}\equiv5^{-1}\pmod7[/tex] is the number L such that

[tex]5L\equiv1\pmod7[/tex]

Consider the first 7 multiples of 5:

5, 10, 15, 20, 25, 30, 35

Taken mod 7, these are equivalent to

5, 3, 1, 6, 4, 2, 0

This tells us that 3 is the inverse of 5 mod 7, so L = 3.

Similarly, compute the inverses modulo 7 of 2 and 3:

[tex]2a\equiv1\pmod7\implies a\equiv4\pmod7[/tex]

since 2*4 = 8, whose residue is 1 mod 7;

[tex]3b\equiv1\pmod7\implies b\equiv5\pmod7[/tex]

which we got for free by finding the inverse of 5 earlier. So

[tex]2^{-1}+3^{-1}\equiv4+5\equiv9\equiv2\pmod7[/tex]

and so R = 2.

Then L - R = 1.