Can anybody suggest a simple proof that shows that (210n )-1 is always divisible by 11 for integer n> 0.
Well, it's true when n=1; 2^10-1=1023=11*93
Assume it's true for x=2^(10k)-1
2^(10(k+1))-1 = (x+1)(2^10)-1 = (x+1)(1024)-1 =
1024x+1023
1024x is divisible by 11 because x is divisible by 11;
1023 is divisible by 11
so 1024x+1023 is divisible by 11
so 2^(10(k+1))-1 is divisible by 11
another way is via modular arithmetic:
210n-1=322n-1 º 102n-1 (mod 11) º (-1)2n-1 (mod 11) º 1n-1 (mod 11) º 0 (mod 11)Another way is to observe that:
an -1=(a-1)(an-1 +...+1)
Substitute
a=210
and noting that 210 -1 is divisible by 11 and you are
done.
Andre
Just to add yet another solution,
there's a result called Fermat's little theorem that implies that
a10 - 1 is always divisible by 11 if a isn't a
multiple of 11. Set a = 2n and you're done.
David
Just for the avoidance of doubt, David's proof is dependent
upon 11 being a prime number.
Andre
Many thanks to everyone who contributed to this thread. I did
not expect so many interesting and varied solutions to this
problem.
John