Let $a\equiv (3^{-1}+5^{-1}+7^{-1})^{-1}\pmod{11}$. What is the remainder when $a$ is divided by $11$?
(3 inverse + 5 inverse + 7 inverse) inverse = (4 + 9 + 2) inerse
= 15 inverse
= 3
Thank you, but the corrrect answer is 10.
Can you try this problem?
What is the smallest integer n, greater than 1, such that n^-1 is defined?
