Cycle Length of the Positive Powers of Two Mod Powers of Ten

  1. 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.
     
  2. jcsd
  3. That was an example for m=1. I should have said, 1 x 4*5^(m-1).
     
  4. 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.
     
  5. 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)
    I see it in examples, but not algebraically. Could you elaborate?
     
  6. 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.
     
  7. 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).

    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)?
     
  8. 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.
     
  9. 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!
     
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