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January 1st, 2019, 08:51 AM   #1
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Divisibility proof

How to show that $\displaystyle \; 11^{n} -6\; $ is divisible by $\displaystyle 5$
Without induction
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January 1st, 2019, 08:55 AM   #2
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11 = 1 mod 5
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January 1st, 2019, 09:07 AM   #3
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Just posted to work without induction

Last edited by idontknow; January 1st, 2019 at 09:09 AM.
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January 1st, 2019, 09:57 AM   #4
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Quote:
Originally Posted by idontknow View Post
How to show that $\displaystyle \; 11^{n} -6\; $ is divisible by $\displaystyle 5$
Without induction
This seems to only work for odd n...

-Dan
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January 1st, 2019, 10:34 AM   #5
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Quote:
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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.
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January 1st, 2019, 11:13 AM   #6
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$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} = $

$0 \pmod{5}$
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January 1st, 2019, 12:47 PM   #7
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Pffl!!

I was using 11n, not $\displaystyle 11^n$. Sorry about that.

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January 1st, 2019, 06:00 PM   #8
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Quote:
Originally Posted by Micrm@ss View Post
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.
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January 1st, 2019, 07:09 PM   #9
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$\displaystyle (5+6)^n-6\rightarrow5^n\cdots6^n-6\rightarrow5^n-6(6^{n-1}-1)$

Six to any positive power is a number with six as the last digit. I will leave it to the reader to finish up.
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Last edited by greg1313; January 2nd, 2019 at 12:58 AM.
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January 2nd, 2019, 02:33 AM   #10
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Of course, one could apply the same sort of idea to $11^n$ ...
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