Question

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 $8$.

Answer 1

These integers are the ones less than and relatively prime to 8, belonging to the set {1, 3, 5, 7}. They happen to be their own inverses (mod 8), since

`1\cdot1 \equiv 1 \pmod 8 \implies 1^(-1) \equiv 1 \pmod 8`

`3\cdot3 \equiv 9 \equiv 1 \pmod 8 \implies 3^(-1) \equiv 3 \pmod 8`

`5\cdot5 \equiv 25 \equiv 1 \pmod 8 \implies 5^(-1) \equiv 5 \pmod 8`

`7\cdot7 \equiv 49 \equiv 1 \pmod 8 \implies 7^(-1) \equiv 7 \pmod 8`

Without loss of generality, let
`\{a,b,c,d\}` be these integers, in that order. Then

`abc \equiv 15 \equiv 7 \pmod 8`

`abd \equiv 21 \equiv 5 \pmod 8`

`acd \equiv 35 \equiv 3 \pmod 8`

`bcd \equiv 105 \equiv 1 \pmod 8`

`abcd \equiv 105 \equiv 1 \pmod 8 \implies (abcd)^(-1) \equiv 1 \pmod 8`

So, we have

`(abc + abd + acd + bcd) (abcd)^(-1) \equiv (7 + 5 + 3 + 1) \cdot 1 \equiv 16 \equiv \boxed{0} \pmod 8`