Math Is Fun Forum

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Proof

EDIT: Never mind.

Last edited by Daniel123 (2008-09-30 03:01:44)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Proof

is the sum of the GP

– which is of course clearly always an integer for

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Proof

Yep. I realised that

could be rewritten as

as soon as I submitted the post, but I had to rush out for a driving lesson.

Last edited by Daniel123 (2008-09-30 04:40:44)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Proof

That's a rather slick algebraic way to see if.  If you didn't see that trick, then you could just brute force it by induction.  You wind up with looking at properties modulo 3, and from there it should be rather immediate.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Proof

you could also do like this:

since exactly one of the three consecutive integers 2^n-1,2^n,2^n+1 must be divisible by 3, and 2^n isnt, 3|(2^n-1) or 3|(2^n+1)->3|(2^n-1)(2^n+1)

Last edited by Kurre (2008-09-30 08:12:40)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Proof

In general,

divides

for all positive integers a and n.

Kurre, your method won’t work for the general case. (Or, more precisely, in cases when

is even.)

Just think sum of GP. That will always work.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Proof

I like Ricky's way.

4^n = 1^n = 1 (mod 3)
Hence, 4^n - 1 = 0 (mod 3) and so will be divisible by 3 for any n.

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Why did the vector cross the road?
It wanted to be normal.