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 $m=11$. Let $L$ be the residue of $(2+3)^{-1}\pmod{11}$, and let $R$ be the residue of $2^{-1}+3^{-1}\pmod{11}$, where $L$ and $R$ are integers from $0$ to $10$ (inclusive). Find $L-R$.
We know that
\begin{align*} (2 + 3) (2^{-1} + 3^{-1}) &\equiv 1 \pmod{11} \\ 2 \cdot 2^{-1} + 2 \cdot 3^{-1} + 3 \cdot 2^{-1} + 3 \cdot 3^{-1} &\equiv 1 \pmod{11}, \end{align*}
or
\[5 \cdot (2^{-1} + 3^{-1}) \equiv 1 \pmod{11}.\]Therefore, the inverse of 5 modulo 11 is $2^{-1} + 3^{-1},$ which means $(a + b)^{-1} \equiv a^{-1} + b^{-1} \pmod{11}.$ Hence, $L - R = \boxed{0}.$
To find $L$, we need to compute $(2+3)^{-1} \pmod{11}$.
We know that $(a+b)^{-1}$ is congruent to $a^{-1}+b^{-1}$ (modulo $m$), so we can compute $2^{-1}+3^{-1} \pmod{11}$ as follows:
$2^{-1} \pmod{11} = 6$ because $2 \cdot 6 \equiv 1 \pmod{11}$.
$3^{-1} \pmod{11} = 4$ because $3 \cdot 4 \equiv 1 \pmod{11}$.
Therefore, $2^{-1}+3^{-1} \pmod{11} = 6+4 \pmod{11} = 10 \pmod{11}$.
Thus, $R=10$.
Now, let's compute $(2+3)^{-1} \pmod{11}$.
$2+3 = 5$, so we need to find the inverse of $5 \pmod{11}$.
To find the inverse of $5 \pmod{11}$, we can try multiplying $5$ by different residues modulo $11$ until we get $1$:
$5 \cdot 2 \equiv 10 \pmod{11}$.
$5 \cdot 3 \equiv 8 \pmod{11}$.
$5 \cdot 4 \equiv 1 \pmod{11}$.
Therefore, $(2+3)^{-1} \pmod{11} = 4$.
Thus, $L=4$.
Finally, we need to find $L-R$:
$L-R = 4 - 10 = -6$.
However, since the residues are taken modulo $11$, we can rewrite $-6$ as $-6 + 11 = 5$.
Therefore, $L-R = \boxed{5}$.
To solve this problem, we need to calculate the values of $(2+3)^{-1} \pmod{11}$ and $2^{-1} + 3^{-1} \pmod{11}$.
First, let's calculate $(2+3)^{-1} \pmod{11}$. To find the inverse of $b \pmod{m}$, we need to find an integer $b^{-1}$ such that $bb^{-1} \equiv 1 \pmod{m}$. In this case, we are looking for an integer $(2+3)^{-1}$ such that $(2+3)(2+3)^{-1} \equiv 1 \pmod{11}$. Let's check each residue from $0$ to $10$ to find the value of $(2+3)^{-1}$:
- If we choose $L=0$ as the residue, then $(2+3)(0) \equiv 0 \not\equiv 1 \pmod{11}$.
- If we choose $L=1$ as the residue, then $(2+3)(1) \equiv 5 \not\equiv 1 \pmod{11}$.
- If we choose $L=2$ as the residue, then $(2+3)(2) \equiv 10 \not\equiv 1 \pmod{11}$.
- If we choose $L=3$ as the residue, then $(2+3)(3) \equiv 15 \equiv 4 \not\equiv 1 \pmod{11}$.
- If we choose $L=4$ as the residue, then $(2+3)(4) \equiv 20 \equiv 9 \not\equiv 1 \pmod{11}$.
- If we choose $L=5$ as the residue, then $(2+3)(5) \equiv 25 \equiv 3 \not\equiv 1 \pmod{11}$.
- If we choose $L=6$ as the residue, then $(2+3)(6) \equiv 30 \equiv 8 \not\equiv 1 \pmod{11}$.
- If we choose $L=7$ as the residue, then $(2+3)(7) \equiv 35 \equiv 2 \not\equiv 1 \pmod{11}$.
- If we choose $L=8$ as the residue, then $(2+3)(8) \equiv 40 \equiv 7 \not\equiv 1 \pmod{11}$.
- If we choose $L=9$ as the residue, then $(2+3)(9) \equiv 45 \equiv 1 \pmod{11}$. Therefore, $L=9$ is the residue of $(2+3)^{-1} \pmod{11}$.
Next, let's calculate $2^{-1} + 3^{-1} \pmod{11}$. We need to find the sum of the inverse of $2$ and the inverse of $3$ modulo $11$. Using the same approach as before, let's check each residue from $0$ to $10$ to find the sum:
- If we choose $R=0$ as the residue, then $2^{-1} + 3^{-1} \equiv 0 \not\equiv 9 \pmod{11}$.
- If we choose $R=1$ as the residue, then $2^{-1} + 3^{-1} \equiv 1 \not\equiv 9 \pmod{11}$.
- If we choose $R=2$ as the residue, then $2^{-1} + 3^{-1} \equiv 5 \not\equiv 9 \pmod{11}$.
- If we choose $R=3$ as the residue, then $2^{-1} + 3^{-1} \equiv 4 \not\equiv 9 \pmod{11}$.
- If we choose $R=4$ as the residue, then $2^{-1} + 3^{-1} \equiv 9 \equiv 9 \pmod{11}$. Therefore, $R=4$ is the residue of $2^{-1} + 3^{-1} \pmod{11}$.
Finally, we calculate $L - R = 9 - 4 = 5$. Therefore, the value of $L-R$ is $5$.