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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 Thanks from idontknow
 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
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Quote:
 Originally Posted by idontknow How to show that $\displaystyle \; 11^{n} -6\;$ is divisible by $\displaystyle 5$ Without induction
This seems to only work for odd n...

-Dan

January 1st, 2019, 11:34 AM   #5
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Quote:
 Originally Posted by topsquark This seems to only work for odd n... -Dan
It doesn't. It works for all n. Can you check again and perhaps post why you don't think it works? I'm curious whether I did something wrong.

 January 1st, 2019, 12:13 PM #6 Senior Member     Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 $11^n - 6 \pmod{5} =$ $(2\cdot 5 + 1)^n - 6 \pmod{5} =$ $\left(\sum \limits_{k=0}^n~\dbinom{n}{k}(2\cdot 5)^k\right) - 6 \pmod{5} =$ $1 - 1 \pmod{5} =$ $0 \pmod{5}$ Thanks from idontknow
 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 timey-wimey stuff. Pffl!! I was using 11n, not $\displaystyle 11^n$. Sorry about that. -Dan Thanks from greg1313
January 1st, 2019, 07:00 PM   #8
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Quote:
 Originally Posted by Micrm@ss It works for all n.
No, it works for every non-negative integer $n$.

For such $n$, x - 1 is a factor of the polynomial x$^n$ - 1.

Hence 11$^n$ - 1 is a multiple of 10, and so 11$^n$ - 6 is divisible by 5.

 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 $\displaystyle (5+6)^n-6\rightarrow5^n\cdots6^n-6\rightarrow5^n-6(6^{n-1}-1)$ Six to any positive power is a number with six as the last digit. I will leave it to the reader to finish up. Thanks from topsquark 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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