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- From: brock@NeXTwork.Rose-Hulman.Edu (Bradley W. Brock)
- Subject: Re: n doesnt divide 2^n-1
- Message-ID: <1992Sep7.200335.3204@cs.rose-hulman.edu>
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- References: <Sep.7.13.02.21.1992.5036@pepper.rutgers.edu>
- Date: Mon, 7 Sep 1992 20:03:35 GMT
- Lines: 19
-
-
- In article <Sep.7.13.02.21.1992.5036@pepper.rutgers.edu>
- gore@pepper.rutgers.edu (Bittu) writes:
- > Conj: For n >= 1, let 2^n = a.n + b where 0 <= b < n so that a is the
- > quotient and b the remainder. Then the following hold:
- >
- > 1. b is always even.
- > 2. (even stronger) b is always a power of 2.
- > 3. a is always even.
- >
- > Note that 1 implies that n doesn't divide 2^n - 1 for n > 1. 2 gives
- > us a slightly stronger theorem though I don't know how to say it. I
- > don't know what 3 implies. My question is: Is any of the three
- > statements true?
- If n=25, a= 1342177 and b=7. Hence, all three statements are false.
- --
- Bradley W. Brock, Department of Mathematics
- Rose-Hulman Institute of Technology | "Honor one another above yourselves."
- brock@nextwork.rose-hulman.edu | -Paul of Tarsus to the Romans
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