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Re: [PrimeNumbers] Powers of Two Primes

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  • Peter Kosinar
    ... Proof: Sum(i=0..N, 10^something(i)*2^i) mod 3 = Sum(i=0..N, 2^i) mod 3 = 2^(N+1)-1 mod 3 = 0 for odd N, 1 for even N. Q.E.D. Peter -- [Name] Peter Kosinar
    Message 1 of 2 , Aug 11, 2007
      > The other conjecture is that alternate power of two numbers
      > are divisible by 3.

      Proof:
      Sum(i=0..N, 10^something(i)*2^i) mod 3 =
      Sum(i=0..N, 2^i) mod 3 =
      2^(N+1)-1 mod 3 =
      0 for odd N, 1 for even N.
      Q.E.D.

      Peter

      --
      [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
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