There are four distinct positive integers $a,b,c,d$ less than $8$ which are invertible modulo $8$. Find the remainder when $(abc+abd+acd+bcd)(abcd)^{-1}$ is divided by $8$.
To find the four distinct positive integers less than $8$ that are invertible modulo $8$, I need to recall what it means for an integer to be invertible modulo $8$. The remainders when dividing any integer by $8$ are $0, 1, 2, 3, 4, 5, 6, 7$. Out of these, only $1, 3, 5, 7$ are coprime to $8$, so they are the only possible values for $a,b,c,d$. Since they are distinct, I can assume without loss of generality that $a = 1, b = 3, c = 5, d = 7$. Then the expression $(abc+abd+acd+bcd)(abcd)^{-1}$ becomes $(15+21+35+105)(105)^{-1}$. 