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May 2nd, 2017, 07:42 PM  #1 
Newbie Joined: May 2017 From: Shanghai, China Posts: 1 Thanks: 0  Divisibility Rules? https://brilliant.org/discussions/th...ilityrules4/ Prove that if x is divisible by 3, then $\displaystyle 2^x  1$ is divisible by 7 
May 3rd, 2017, 12:20 AM  #2  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Quote:
Then $2^x  1 = 2^{3y}  1$ $2^{3y}  1 = (2^3)^y  1$ Factor $2^{3y}  1 = (2^3  1)[(2^3)^{y 1} + (2^3)^{y2} + ... + 1)]$ $2^{3y}  1 = 7k$ This shows $ \ \ \ 2^x  1 \ \ $ (given your condition) is a multiple of $ \ 7 \ $ therefore divisible by $ \ 7$ Last edited by agentredlum; May 3rd, 2017 at 12:22 AM.  
May 3rd, 2017, 12:22 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,140 Thanks: 1415 
The polynomial $y^n  1$ is divisible by $y  1$ by the factor theorem. Hence if $y = a^b$ for positive integers $a$ and $b$, $a^b  1$ divides $y^n  1$, which is $a^{bn}  1$. For the given problem, use $a = 2$ and $b = 3$ so that $a^b 1 = 7$. 

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