Cube conjecture
Ed Pegg Jr.
Is there an easy way to disprove this conjecture? Is there a sequence of
numbers whose cubes never contain a zero?
Another similar conjectures:
Only finitely many squares have no internal doubled digits.
--Ed Pegg Jr, www.mathpuzzle.com
Domnei
>
>Is there an easy way to disprove this conjecture? Is there a sequence of
>numbers whose cubes never contain a zero?
>
>Another similar conjectures:
>
>Only finitely many squares have no internal doubled digits.
>
Ed -
I believe these kind of conjectures are generally very
hard (read, impossible, so far!) to prove. Others I can
think of along the same lines, that I've either looked into
a bit or have come up in problems I've worked on, include:
- only finitely many nth powers lack the digit zero (for
various n's)
- only finitely many powers of 2 (or powers of 3, etc.)
lack the digit zero
- only finitely many numbers of the form (2^a)(3^b)(5^c)(7^d)
lack the digit zero
and so on. As far as I know, all of these are unproven
(but conjectured to be true). I didn't just make up the 3rd
one...if true, it would settle "Conjecture 1" in
http://users.aol.com/s6sj7gt/mari.htm
Mike Keith
Web site: http://users.aol.com/s6sj7gt/mikehome.htm
(remove post-w letters from e-mail address)
Ed Pegg Jr.
> Do you mean 10^3 , 100^3 , 1000^3.. etc.?
No, those numbers contain the digit zero.
137^3 = 2571353 lacks a zero. The conjecture involves cubes containing
billions of digits. Using a bit of probability, I feel it's a safe
conjecture.
> |Only finitely many cubic numbers lack the digit zero.
> |--Ed Pegg Jr, www.mathpuzzle.com
Tapio Hurme
Ed Pegg Jr. wrote in message <7fdehs$6...@dfw-ixnews10.ix.netcom.com>...
Dean Hickerson
: Only finitely many cubic numbers lack the digit zero.
: Is there an easy way to disprove this conjecture? Is there a sequence of
: numbers whose cubes never contain a zero?
I looked at this about 10 years ago, and found such a sequence:
If n == 2 (mod 3) and n >= 5, then the cube of
5n 4n 3n-1 2n n
2*10 - 10 + 17*10 + 10 + 10 - 2
N = ------------------------------------------
3
lacks the digit 0. For example, with n = 23,
N = 666666666666666666666663333333333333333333333390000000000000000000000\
3333333333333333333333366666666666666666666666
and
3
N = 296296296296296296296291851851851851851851851949629629629629629629629\
281481481481481481481489792592592592592592592513814814814814814814816\
596777777777777777777763211111111111111111111148777777777777777777776\
994814814814814814814823592592592592592592592461481481481481481481481\
829629629629629629629631851851851851851851851896296296296296296296296
Dean Hickerson
de...@math.ucdavis.edu
Clive Tooth
>I believe these kind of conjectures are generally very
>hard (read, impossible, so far!) to prove. Others I can
>think of along the same lines, that I've either looked into
>a bit or have come up in problems I've worked on, include:
>
>- only finitely many nth powers lack the digit zero (for
> various n's)
>- only finitely many powers of 2 (or powers of 3, etc.)
> lack the digit zero
>- only finitely many numbers of the form (2^a)(3^b)(5^c)(7^d)
> lack the digit zero
>
>and so on. As far as I know, all of these are unproven
>(but conjectured to be true). I didn't just make up the 3rd
>one...if true, it would settle "Conjecture 1" in
>
>http://users.aol.com/s6sj7gt/mari.htm
Something that I once proved:
There exist powers of 2 containing arbitrarily long strings of consecutive
zeros in their decimal forms.
Clive
Main Night
>zeros in their decimal forms.
Show! Show! Show!
Peace out
Sam =)
Robert Israel
Main Night <main...@aol.com> wrote:
>>There exist powers of 2 containing arbitrarily long strings of consecutive
>>zeros in their decimal forms.
>
>Show! Show! Show!
Hint: log_10(2) is irrational. A power of 2 can start with an
arbitrary string of digits (except of course that the first is nonzero).
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
ni...@mathematik.tu-muenchen.de
(and subsequently lost the temp file). :^(
main...@aol.com (Main Night) wrote:
> References: <7ff4jr$q1b$1...@mail.pl.unisys.com>
(Moreover it doesn't seem to want to put that into the headers
where it belongs.)
>>There exist powers of 2 containing arbitrarily long strings of consecutive
>>zeros in their decimal forms.
>
>Show! Show! Show!
2^70777 = 1000007160137454597936006638....124245908692653700022272
^^^^^
because 70777*(log 2)/(log 10) is just a trifle larger than 21306.
Some background: (log 2)/(log 10) is irrational (easy consequence
of unique factorization in the integers).
Therefore (well - by a Dirichlet pigeon-hole argument, one of the
easiest results of the discipline known as Diophantine Approximation),
given any epsilon > 0, there exist infinitely many k such that the
fractional part of (log 2)/(log 10) is (positive and) less than
epsilon. Using such a k, we find that
2^k = 10^(k*(log 2)/(log 10))
equals an integral power of 10 times a number larger than 1 and smaller
than 10^epsilon. If epsilon was small enough, this means the leading
digits will be a 1 followed by any desired number of zeros.
This generalizes to the statement that we can obtain any desired finite
string of digits infinitely often as the leading digits of powers of 2,
written in any base which isn't itself a power of 2. And further to
powers of 3,...
If you want to find out how I discovered the 70777, rather quicker and
more elegantly than by a brute-force search, you'll want to learn about
continued fractions. Hardy's & Wright's `An introduction to the theory
of numbers' is probably the most useful reference.
Enjoy, Gerhard
--
* Gerhard Niklasch <ni...@mathematik.tu-muenchen.de> * spam totally unwelcome
* http://hasse.mathematik.tu-muenchen.de/~nikl/ ******* all browsers welcome
* This .signature now fits into 3 lines and 77 columns * newsreaders welcome
Andres Koropecki
isr...@math.ubc.ca (Robert Israel) wrote:
> In article <19990419161601...@ng114.aol.com>,
> Main Night <main...@aol.com> wrote:
> >>zeros in their decimal forms.
> >
> >Show! Show! Show!
>
> arbitrary string of digits (except of course that the first is nonzero).
>
> Robert Israel isr...@math.ubc.ca
As a matter of fact, there is power of n which starts with any arbitrary
string of digits for every positive integer n, exeptuating those n's which
are powers of 10. You can find a proof of this in Ross Honsberger
"Ingenuity in Mathematics". (Ch. 6)
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http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
dr_a...@my-dejanews.com
ni...@mathematik.tu-muenchen.de wrote:
> Rats, Lynx seems to have mangled my first attempt at a follow-up
> (and subsequently lost the temp file). :^(
>
> main...@aol.com (Main Night) wrote:
> > References: <7ff4jr$q1b$1...@mail.pl.unisys.com>
>
> (Moreover it doesn't seem to want to put that into the headers
> where it belongs.)
>
> >>zeros in their decimal forms.
> >
> >Show! Show! Show!
>
Yea Hardy and Wright is right and hardy for number theory, but for more
specialized questions like this, i feel Sierpinski "250 problems in number
theory" or "Elementary number theory" is really cool.
For instance, in Sierpinski's book(s) he shows :
given any sequence of numbers, it is possible to append the sequence on the
right to make it into a perfect power of 2.
From which of course the comment about there existing powers of 2's with
long trails of 0's follows trivially (funnier perhaps is the fact that
there is a power of 2 which starts out as 314...., one which starts out
as 314159265.....)
Cheers,
Sam.
Clive Tooth
The smallest power of two to contain a run of 13 consecutive zeros is
2^4036339 .
Clive
Main Night
>The smallest power of two to contain a run of 13 consecutive zeros is
>2^4036339 .
1) Show me the method used to prove this
2) Show me the computer which runs the method
3) Define the computer as a mathematical object and prove to me that it didn't
make an error in the computation
Peace out
Sam =)
Domnei
>didn't
>make an error in the computation
>
Oh, please.
Main Night
2000 years ago.......
"Mr. Pythagoras, that is a truly beautiful result!"
"Thank you!"
"But, Mr. Pythagoras, could you kindly demonstrate that there are no
incommensureable magnitudes?"
"Oh, please."
Peace out
Sam =)
Iain Davidson
In article <7fg66f$5c5$1...@nntp.ucs.ubc.ca>, Robert Israel (isr...@math.ubc.ca) writes:
>In article <19990419161601...@ng114.aol.com>,
>Main Night <main...@aol.com> wrote:
>>>zeros in their decimal forms.
>>
>>Show! Show! Show!
>
>arbitrary string of digits (except of course that the first is nonzero).
Do you mean that if
2^k = 10^l
then k*log(2) = l
log(2) = l/k
If you expand l/k as a continued fraction
(0/1), (1/3),(3/10),(28/93) etc .then you eventually get 2^k = 9**,
10**, 99**, 100** and so on.
Iain Davidson Tel : +44 1228 49944
4 Carliol Close Fax : +44 1228 810183
Carlisle Email : ia...@stt.win-uk.net
England
CA1 2QP