Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2008-11-02 06:52:22

Daniel123
Member
Registered: 2007-05-23
Posts: 663

proof involving base b

Sorry for all the questions! I'm pretty much there on most of them, I just need to iron out a few problems.

My problem here is in part b. I can only show the result if k is even, not if k is odd. Can someone please tell me why? Thanks.

Last edited by Daniel123 (2008-11-02 06:55:52)

Offline

#2 2008-11-02 07:46:53

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: proof involving base b

Maybe I'm misinterpreting, but isn't b) false?

Working in base 10, I think 101 would be an example of the type of number b) is describing.
But 101 isn't divisible by 11, so something's gone wrong.


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

Offline

#3 2008-11-02 07:54:55

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: proof involving base b

Working in base 10,

101 is not divisible by 11, and k is odd

12021 is divisible by 11, and k is even

1230123 is not divisible by 11, and k is odd

473104731 is divisible by 11, and k is even.

Maybe I'm right then? And the question should specify that k is even? (as I can't see anything wrong with my working)

Last edited by Daniel123 (2008-11-02 07:56:01)

Offline

#4 2008-11-02 08:06:17

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: proof involving base b

Looks good to me too.

Just for the sake of clarity, it would be a good idea to write out f(x) before computing f(b).


"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..."

Offline

#5 2008-11-02 08:15:03

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: proof involving base b

Great, thanks smile

Offline

Board footer

Powered by FluxBB