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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 \(n\).)

 Sep 13, 2020
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mmi = Modular Multiplicative Inverse of 1 to 4! mod 4!.

1  -  The mmi =  1
5  -  The mmi =  5
7  -  The mmi =  7
11  -  The mmi =  11
13  -  The mmi =  13
17  -  The mmi =  17
19  -  The mmi =  19
23  -  The mmi =  23

 

m =1 x 5 x 7 x 11 x 13 x 17 x 19 x 23 =37,182,145
37,182,145 mod 4! = 1

 Sep 13, 2020

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