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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.

 Dec 11, 2018
 #1
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\(\text{Let }S \text{ be the set of all positive integers invertible mod }4!\\ \text{you can partition }S \text{ into pairs of inverses }(x,~x^{-1})\\ -\\ \text{The product of all the elements mod 4! is equal to the product of the }\\ \text{products of the individual pairs mod 4!}\\ -\\ x \cdot x^{-1} \pmod {4!} = 1 \text{ by definition of multiplicative inverse}\\ \text{The product of all these 1's will be 1, and thus the overall product mod 4!}=1\)

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 Dec 11, 2018
edited by Rom  Dec 11, 2018

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