Divisibility proof

Dec 2015
1,076
166
Earth
How to show that \(\displaystyle \; 11^{n} -6\; \) is divisible by \(\displaystyle 5\)
Without induction
 
Dec 2015
1,076
166
Earth
Just posted to work without induction
 
Last edited:

topsquark

Math Team
May 2013
2,510
1,049
The Astral plane
How to show that \(\displaystyle \; 11^{n} -6\; \) is divisible by \(\displaystyle 5\)
Without induction
This seems to only work for odd n...

-Dan
 

romsek

Math Team
Sep 2015
2,958
1,673
USA
$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} = $

$0 \pmod{5}$
 
  • Like
Reactions: 1 person

topsquark

Math Team
May 2013
2,510
1,049
The Astral plane
Pffl!! :furious:

I was using 11n, not \(\displaystyle 11^n\). Sorry about that. :cry:

-Dan
 
  • Like
Reactions: 1 person

skipjack

Forum Staff
Dec 2006
21,473
2,466
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.
 
  • Like
Reactions: 1 person

greg1313

Forum Staff
Oct 2008
8,008
1,174
London, Ontario, Canada - The Forest City
\(\displaystyle (5+6)^n-6\rightarrow5^n\cdots6^n-6\rightarrow5^n-6(6^{n-1}-1)\)

Six to any positive power is a number with six as the last digit. I will leave it to the reader to finish up.
 
Last edited:
  • Like
Reactions: 1 person

greg1313

Forum Staff
Oct 2008
8,008
1,174
London, Ontario, Canada - The Forest City
Of course, one could apply the same sort of idea to $11^n$ ...