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MathGroup Archive 2001

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Two factors of (10^71-1)/9 = R71

  • To: mathgroup at smc.vnet.net
  • Subject: [mg30125] Two factors of (10^71-1)/9 = R71
  • From: seidovzf at yahoo.com (Zakir F. Seidov)
  • Date: Sun, 29 Jul 2001 21:26:17 -0400 (EDT)
  • Organization: The Math Forum
  • References: <6kajko$2bn@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

The number William Mopppett  wrote:
           ================ 
(10^71 - 1)/9 
is R71, repunit, number with "all ones".

And it has two factors:
R71 = 
241573142393627673576957439049
*45994811347886846310221728895223034301839

See, e.g.,  
http://www.ping.be/~ping6758/repunits.htm

But suprisingly enough,
when,  before looking for "repunit"s in 37.com,
I asked my PC to work for saturday, 
its Mathematica session was:

 $Version
 4.0 for Microsoft Windows (December 5, 1999)

<< NumberTheory`FactorIntegerECM`

Timing[f = FactorIntegerECM[(2^128 + 1)/f] ]
{793.95 Second, 59649589127497217} (* good, but...*)

Timing[FactorIntegerECM[(10^71 - 1)/9 ]]
{112642. Second, \
111111111111111111111111111111111111111111111111111111111111111\
11111111} (* ??!! *)

That is Mathematica (or my PC?) considers R71 as prime!

And this is a real suprise 
and hence the question to Mathematica experts:
how it can be?

Zakir F. Seidov
seidovzf at yahoo.com

%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%
Subject: [mg30125]      RE: factor
Author:       Ersek_Ted%PAX1A at mr.nawcad.navy.mil
Organization: Steven M. Christensen and Associates, Inc and
MathTensor, Inc.


William Mopppett  wrote:
|
|Can anyone give me a factor of (10^71 - 1)/9 Mathematica 3.0 says it
is
|not prime email
|wmoppet at nsw.bigpond.net.au
|
|

  The attempt below is probably the most effective way to do this. My
90
Mhz Pentium worked on it for 14 hours and still didn't find a factor.

The documentation makes it clear you will may not get an answer in a
short  amount of time.  It also says it is designed to find factors up
to about 18  digits in 3 hours on a "workstation" (what ever that is).

Well it seems all the prime factors of this number have 18 to 35
digits,
and  it has two, three, or four prime factors.  The number of
potential
prime  factors to consider is roughly,
(  10^35/Log[10^35] - 10^18/Log[10^18] ) = ( 1.2 * 10^33  )     !!


In[1]:=
<<NumberTheory`FactorIntegerECM`

In[2]:=
FactorIntegerECM[(10^71 - 1)/9 ]

(*  still waiting  *)
___________________________________
Perhaps you can do better by specifying a value for one or more of the

options.

In[3]:=
Options[FactorIntegerECM]

Out[3]=
{FactorSize -> Automatic,
  CurveNumber -> Automatic,
  CurveCountLimit -> 10000}

__________________________________

Ted Ersek



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