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January 1st, 2019, 09:51 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 327 Thanks: 42  Divisibility proof
How to show that $\displaystyle \; 11^{n} 6\; $ is divisible by $\displaystyle 5$ Without induction 
January 1st, 2019, 09:55 AM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 535 Thanks: 306 Math Focus: Dynamical systems, analytic function theory, numerics 
11 = 1 mod 5

January 1st, 2019, 10:07 AM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 327 Thanks: 42 
Just posted to work without induction
Last edited by idontknow; January 1st, 2019 at 10:09 AM. 
January 1st, 2019, 10:57 AM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,980 Thanks: 789 Math Focus: Wibbly wobbly timeywimey stuff.  
January 1st, 2019, 11:34 AM  #5 
Senior Member Joined: Oct 2009 Posts: 696 Thanks: 235  
January 1st, 2019, 12:13 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 
$11^n  6 \pmod{5} = $ $(2\cdot 5 + 1)^n  6 \pmod{5} = $ $\left(\sum \limits_{k=0}^n~\dbinom{n}{k}(2\cdot 5)^k\right)  6 \pmod{5} = $ $1  1 \pmod{5} = $ $0 \pmod{5}$ 
January 1st, 2019, 01:47 PM  #7 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,980 Thanks: 789 Math Focus: Wibbly wobbly timeywimey stuff. 
Pffl!! I was using 11n, not $\displaystyle 11^n$. Sorry about that. Dan 
January 1st, 2019, 07:00 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,105 Thanks: 1907  
January 1st, 2019, 08:09 PM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond 
$\displaystyle (5+6)^n6\rightarrow5^n\cdots6^n6\rightarrow5^n6(6^{n1}1)$ Six to any positive power is a number with six as the last digit. I will leave it to the reader to finish up. Last edited by greg1313; January 2nd, 2019 at 01:58 AM. 
January 2nd, 2019, 03:33 AM  #10 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond 
Of course, one could apply the same sort of idea to $11^n$ ...


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