### Instruction: Let $a\equiv (3^{-1}+5^{-1}+7^{-1})^{-1}\pmod{11}$. What is the remainder when $a$ is divided by $11$?### Response: I know that the inverse of a number modulo $11$ is another number between $0$ and $10$ such that their product is congruent to $1$ modulo $11$.
For example, $3^{-1}\equiv 4\pmod{11}$, because $3\times 4\equiv 12\equiv 1\pmod{11}$.
Similarly, $5^{-1}\equiv 9\pmod{11}$ and $7^{-1}\equiv 8\pmod{11}$.
So, I can rewrite the expression for $a$ as $a\equiv (4+9+8)^{-1}\pmod{11}$.
Now, I just need to simplify the parentheses and find the inverse of the result.
I know that $4+9+8=21$, which is congruent to $10$ modulo $11$.
So, $a\equiv (10)^{-1}\pmod{11}$.
To find the inverse of $10$ modulo $11$, I can try some trial and error.
I notice that $10\times 10=100$, which is congruent to $1$ modulo $11$.
Therefore, $10^{-1}\equiv 10\pmod{11}$.
So, $a\equiv 10\pmod{11}$.