[math-fun] Integers with many 0s in base 2 and base 3 (fwd)
---------- Forwarded message ---------- Date: Sat, 21 Dec 2002 00:01:59 -0500 From: Andreas Weingartner <weingartner@suu.edu> To: NMBRTHRY@LISTSERV.NODAK.EDU Subject: Integers with many 0s in base 2 and base 3 Hello everybody: Here is a problem that comes up in quantum computation. I would appreciate any information on this problem. Prove or disprove: There exists an epsilon>0, such that no natural number has the property that in base 2 as well as in base 3, at most (epsilon)*100% of the digits are nonzero. Example: 513=(1000000001) in base 2 and 513=(201000) in base 3 so epsilon<1/3. Andreas
Prove or disprove: There exists an epsilon>0, such that no natural number has the property that in base 2 as well as in base 3, at most (epsilon)*100% of the digits are nonzero.
Example: 513=(1000000001) in base 2 and 513=(201000) in base 3 so epsilon<1/3.
272629962 = (10000010000000000000011001010) in base 2, 272629962 = (201000000001000000) in base 3 so epsilon < 6/29 1208614932, 2453670144, 17448310278, 22083026976 all have epsilon 1/5 . --Ed Pegg Jr, www.mathpuzzle.com
At 03:47 PM 12/21/02, ed pegg wrote:
Prove or disprove: There exists an epsilon>0, such that no natural number has the property that in base 2 as well as in base 3, at most (epsilon)*100% of the digits are nonzero.
Example: 513=(1000000001) in base 2 and 513=(201000) in base 3 so epsilon<1/3.
272629962 = (10000010000000000000011001010) in base 2, 272629962 = (201000000001000000) in base 3 so epsilon < 6/29
1208614932, 2453670144, 17448310278, 22083026976 all have epsilon 1/5 .
2542645806624 gives 4/21. -- Fred W. Helenius <fredh@ix.netcom.com>
participants (4)
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ed pegg -
Fred W. Helenius -
Helger Lipmaa -
Mike Stay