Let $a\equiv (3^{-1}+5^{-1}+7^{-1})^{-1}\pmod{11}$. What is the remainder when $a$ is divided by $11$?
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, $a\equiv 10\pmod{11}$. This means that the remainder when $a$ is divided by $11$ is $10$.

