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What is the remainder when $10!$ is divided by $2^{7}$? (Reminder: If $n$ is a positive integer, then $n!$ stands for the product $1\cdot 2\cdot 3\cdot \cdots \cdot (n-1)\cdot n$.)

 Dec 24, 2020
 #1
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10!  = 1 * 2 * 3  * (2 * 2)  *  5  * ( 3 * 2)  * 7 * ( 2^3)  * 9  *  (5 * 2)  / 2^7   =

 

2^8   (  3 * 5 * 3 * 7 * 9 * 5)                       

________________________   =          2 ( 3^2 * 5^2 * 7 * 9)   =    28350 

             2^7

 

(Remainder  =  0 )

 

cool cool cool

 Dec 24, 2020
edited by CPhill  Dec 24, 2020
 #2
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+1

10! = [2^8 * 3^4 * 5^2 * 7] mod 2^7 ==0 - The remainder.

 Dec 24, 2020

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