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-   -   Cycle Length of the Positive Powers of Two Mod Powers of Ten (http://www.physicsforums.com/showthread.php?t=347011)

DoctorBinary Oct19-09 11:35 AM

Cycle Length of the Positive Powers of Two Mod Powers of Ten
 
The positive powers of 2 mod 5^m cycle with period 4*5^(m-1), which you can prove by showing that 2 is a primitive root mod powers of 5. I want to prove that the positive powers of two, mod 10^m, also cycle with this same period. How do I go from this powers of 5 result to powers of 10?

The Chinese Remainder Theorem (CRT) obviously came to mind, so I considered the powers of 2 mod 2^m. Starting at 2^m, they cycle with period 1 (they are always 0). The answer, I'm sure you're going to say, is to multiply the two periods: 1 x 4 = 4. But what specifically about the CRT lets me do that? The statements of the CRT I've seen talk about the residues, not the periods.

Maybe I need to invoke an underlying theorem from group theory instead?

Thanks.

DoctorBinary Oct21-09 11:28 AM

Re: Cycle Length of the Positive Powers of Two Mod Powers of Ten
 
Quote:

Quote by DoctorBinary (Post 2400449)
... is to multiply the two periods: 1 x 4 = 4.

That was an example for m=1. I should have said, 1 x 4*5^(m-1).

hamster143 Oct22-09 04:12 AM

Re: Cycle Length of the Positive Powers of Two Mod Powers of Ten
 
Periodicity implies that there's i0 such that, for all i>=i0, 2^(i+p) = 2^i mod 5^m,
therefore 2^(i+p) = 2^i + n*5^m.

It's easy to see that n = n' * 2^i0. If i0 >= m, 2^(i+p) = 2^i + n' * 10^m, which proves that your original period mod 5^m is also a period mod 10^m.


On the other hand, any period mod 10^m must be a period mod 5^m:

2^(i+p) = n * 10^m + 2^i = (n*2^m)*5^m + 2^i


Which proves that two periods are equal.

DoctorBinary Oct23-09 10:52 AM

Re: Cycle Length of the Positive Powers of Two Mod Powers of Ten
 
Nice approach (I guess in a way it's using the CRT implicitly). Thanks.

I have two questions though:

1) Isn't i0 = m?


2)
Quote:

Quote by hamster143 (Post 2405041)
It's easy to see that n = n' * 2^i0

I see it in examples, but not algebraically. Could you elaborate?

hamster143 Oct23-09 04:47 PM

Re: Cycle Length of the Positive Powers of Two Mod Powers of Ten
 
1) Maybe, I wasn't trying to prove that.

2) 2^(i+p) = 2^i + n*5^m, therefore 2^i * (2^p - 1) = n*5^m, RHS is divisible by 2^i.

DoctorBinary Oct26-09 10:49 AM

Re: Cycle Length of the Positive Powers of Two Mod Powers of Ten
 
Quote:

Quote by hamster143 (Post 2407428)
1) Maybe, I wasn't trying to prove that.

I think i0 HAS to be m in order for it to work (to get the 2^m you need to match the 5^m).

Quote:

Quote by hamster143 (Post 2407428)
On the other hand, any period mod 10^m must be a period mod 5^m

I was wondering -- is this direction necessary? Isn't showing that the period mod 5^m is the period mod 10^m sufficient? How could there be more than one period for the powers of two mod 10^m (for any m)?

hamster143 Oct26-09 12:38 PM

Re: Cycle Length of the Positive Powers of Two Mod Powers of Ten
 
i0 has to be m or greater.

If p is the smallest period mod 5^m and we can prove that it is also a period mod 10^m, it's still possible that p is the multiple of the smallest period mod 10^m.

DoctorBinary Oct26-09 01:01 PM

Re: Cycle Length of the Positive Powers of Two Mod Powers of Ten
 
OK, I get it now. The issue is that 10^m's period COULD be less than 5^m's period. The second part of your proof shows it's not.

As for i0, I would still say it's m because your statement "for all i>=i0" itself implies m or greater.

I REALLY appreciate the help!

DoctorBinary Nov9-09 10:59 AM

Re: Cycle Length of the Positive Powers of Two Mod Powers of Ten
 
I wrote this proof up in detail here: http://www.exploringbinary.com/cycle...powers-of-ten/ .


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