Let m be the product of all positive integers less than 4! which are invertible modulo 4!. Find the remainder when m is divided by 4!. (Here n! denotes 1 \(\times \cdots \times\)n for each positive integer n.)

I am not really sure how to go about solving or answering this problem.

OkNietzsche Dec 11, 2018

#1**+1 **

\(\text{Let }S \text{ be the set of all positive integers invertible mod }4!\\ \text{you can partition }S \text{ into pairs of inverses }(x,~x^{-1})\\ -\\ \text{The product of all the elements mod 4! is equal to the product of the }\\ \text{products of the individual pairs mod 4!}\\ -\\ x \cdot x^{-1} \pmod {4!} = 1 \text{ by definition of multiplicative inverse}\\ \text{The product of all these 1's will be 1, and thus the overall product mod 4!}=1\)

.Rom Dec 11, 2018