Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I want to prove that the positive powers of two, mod 10m, cycle with period 4*5m-1. It's simple to prove that the powers of FIVE cycle with this period (2 is a primitive root mod powers of five), but how do you make the leap to powers of TEN?

I'm sure it's something simple -- perhaps related to the Chinese Remainder Theorem -- but I don't see the connection yet.

Thanks for the help.

share|improve this question

closed as off-topic by Ricardo Andrade, Andrey Rekalo, Olivier Benoist, Stefan Kohl, Willie Wong Nov 28 '13 at 12:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, Andrey Rekalo, Olivier Benoist, Stefan Kohl, Willie Wong
If this question can be reworded to fit the rules in the help center, please edit the question.

    
It's exactly the Chinese Remainder Theorem. –  Qiaochu Yuan Oct 17 '09 at 19:19
    
Could you elaborate please? –  Rick Regan Oct 17 '09 at 20:04
    
You know that the powers of two have a certain period mod 5^m. What is their period mod 2^m? –  Qiaochu Yuan Oct 17 '09 at 21:26
    
Their ''period'' would always be 1 (powers are always 0). –  Rick Regan Oct 17 '09 at 22:02
1  
So if M is a number which leaves a residue of 1 mod 5^k, and a residue of 0 mod 2^k, what residue does it leave mod 10^k? –  Alon Amit Oct 17 '09 at 22:59
show 8 more comments

2 Answers 2

And if you insist, let me write this out in detail. All you need is the following lemma.

Lemma: Let f(n) be periodic with period p and let g be injective. Then g(f(n)) is periodic with period p.

Proof. Clearly g(f(n+p)) = g(f(n), so g(f(n)) has some period q dividing p. On the other hand, g(f(n+q)) = g(f(n)) for all n if and only if f(n+q) = f(n) for all n by injectivity, so q = p.

As I remarked above we have bn = b for all but finitely many n and x -> CRT(x, b) is an injection. The result follows.

share|improve this answer
    
Thanks Qiaochu. (I got another answer at physicsforums.com/showthread.php?p=2400449#post2400449 which you might want to check out. It appeals to me more because it uses algebra and exponent arithmetic and does not use the CRT explicitly.) –  Rick Regan Oct 28 '09 at 13:47
add comment
up vote -1 down vote accepted

The answer I like best is based on the proof in the "physics forums" thread linked to in the comments above. I wrote about it in detail here: http://www.exploringbinary.com/cycle-length-of-powers-of-two-mod-powers-of-ten/

share|improve this answer
    
Not sure about the etiquette of accepting your "own " answer... –  Yemon Choi Nov 9 '09 at 20:42
    
From the FAQ: "It's also perfectly fine to ask and answer your own question... Also, we recommend not answering your own question immediately since leaving the question "open" encourages other people to respond to it. You might discover that somebody else has an even better answer!" I think waiting 23 days for other answers was sufficient. Also, I cited a source upon which I based my answer, but it was given in piecemeal fashion in a forum thread. I wrote an article to try to present the answer clearly -- and it was long with lots of math notation -- so I cited it here instead of copying it. –  Rick Regan Nov 9 '09 at 22:29
    
This is also from the FAQ: "you don't earn any reputation for accepting your own answer to a question". So in other words, they're implying it's OK to accept your own answer. –  Rick Regan Nov 9 '09 at 22:32
    
Fair enough. Reverting (i.e. +1) –  Yemon Choi Nov 10 '09 at 18:26
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.