We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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