From: walterhNOSPAM@gmx.de (Walter Hofmann) Subject: Re: Zeros in 2^n Date: Wed, 23 Feb 2000 01:25:30 +0100 Newsgroups: sci.math Summary: [missing] On Mon, 21 Feb 2000 18:45:25 -0500, Jan Kristian Haugland wrote: > >On Mon, 21 Feb 2000, Walter Hofmann wrote: > >> On Wed, 16 Feb 2000 21:59:05 +0100, Andrej wrote: >> > >> > >> >What is the largest n for which the 2^n doesn't have any zeros in >> >decimal form. >> >> Try using the chinese remainder theorem. > >How would that help? Fix a digit. Determine the period of the sequence of this digit for n=1,2,3,... . Do this for a number of digits. Calculate the least common multiple m of all periods you found. Make a list of all congruence classes modulo m for which one of the digits is zero. Hope that for _every_ congruence class mudulo m one of the digits is zero. Then check all powers of two up to 2^i where i is big enough so that all digits you have used for m are within their period. Walter ============================================================================== From: israel@math.ubc.ca (Robert Israel) Subject: Re: Zeros in 2^n Date: 23 Feb 2000 19:21:08 GMT Newsgroups: sci.math In article , walterhNOSPAM@gmx.de (Walter Hofmann) writes: > Fix a digit. Determine the period of the sequence of this digit for > n=1,2,3,... . Do this for a number of digits. Calculate the least common > multiple m of all periods you found. Make a list of all congruence > classes modulo m for which one of the digits is zero. Hope that for > _every_ congruence class mudulo m one of the digits is zero. Then check > all powers of two up to 2^i where i is big enough so that all digits you > have used for m are within their period. Interesting idea. But if 2 is a primitive root of 5^k (I think it is, for all k), the last k digits of 2^n (for n > k) will cycle through 2^k z for all z in {1,...,5^k} not divisible by 5. So the only way this approach would work is if all 4^k of these k-digit integers had at least one zero. Now the probability of a randomly chosen k-digit integer having no zeros is (9/10)^k, so with 4^k of them (admittedly, not randomly chosen) you should expect to have about 3.6^k with no zeros. The conclusion is that it's very unlikely to be true that for any fixed k, all sufficiently large powers of 2 have a zero in their last k digits. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2