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Re: Number theory question continued
Posted:
Mar 27, 2006 4:31 PM


In article <44283c39$1@news.ysu.edu>, Eric J. Wingler <wingler@math.ysu.edu> wrote: > >"ManOfLight" <mladensavov@yahoo.com> wrote in message >news:1143276915.440583.51260@z34g2000cwc.googlegroups.com... >> Hello everybody, >> I post it again to rapair one mistake >> Could you give a clue how I can start the following problem or >> propose a solution. >> >> >> "Is it true that for every sufficiently large interval there will be a >> integer in it of the form >> 2^n3^m where m,n are integers?" >> >> >> As far as I understand it we are supposed either to prove that there >> exists number H : every interval with length H contains such a number >> or disprove it. > >If x = 2^n  3^m, then log(1  x/2^n) = m*log(3)  n*log(2), so it's >possible that you may be able to make use of the density of the set of >numbers of the form m*log(3)  n*log(2).
I doubt it.
This implies there are infinitely many integers of the form x = 2^n  3^m with c < x/2^n < d for any c and d with 0 < c < d < 1. But unfortunately the interval for x grows as n > infty. You can do slightly better using results about Diophantine approximation, but still you won't get an interval of constant length.
Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada



