HDTenEightyP/NetNews huggingface dataset:
From: guckes@math.fu-berlin.de (Sven Guckes)
Subject: Re: primes in 1, 11, 111, etc.
Originator: guckes@leibniz
Organization: Free University of Berlin, Germany
Date: Fri, 17 Apr 1992 14:47:07 GMT

amorgan@Xenon.Stanford.EDU (Crunchy Frog) writes:

>In article <PALMERP.92Apr15220643@ilanga.math.orst.edu> 
>  palmerp@math.orst.edu writes:
>>In article <sa4h9INNrt5@agate.berkeley.edu> 
>>  davidb@triangle.Berkeley.EDU (David T. Blackston) writes:
>>
>>   The question of whether the number of primes consisting of all ones
>>   in their decimal representation is finite is unsolved.  
>>
>>   Here are some interesting (?) related questions.  Let a_n be the 
>>   number consisting of n ones.
>>
>>   Question 1: Show that a_n is prime only if n is prime.  (Easy)
>>
>>I came into this discussion late, so maybe I'm confused. I wrote a 1-line
>>Mathematica program to factor a_n through a_41, and the only primes it found
>>were a_19 and a_23. Here's the program in case someone with Mathematica wanst
>>to check it out. 
>
>.. deleted ..
>
>>Am I just tired/confused?
>
>Confused.  David said a_n is prime *only* if n is prime.  Another way
>of phrasing this is "if a_n is prime, then n is prime"
>
>>Paul Palmer
>>Department of Mathematics     	E-mail: palmerp@math.orst.edu
>
>C Frog

I have found a remark in a book saying

	" I have a complete proof that ... "  (Nono, just joking)

which says that a_n have been tested unto n=1000 and no more primes were
found. It does not give references to who did it when nor how, though.
I think you may feed your Mathematica with other problems now.

Sven


Newsgroups: rec.puzzles
From: ronnie@cisco.com (Ronnie Kon)
Subject: Re: Dying young
Organization: cisco Systems, Incorporated
Date: Thu, 9 Apr 1992 23:42:05 GMT

In article <1992Apr6.175229.23741@hellgate.utah.edu>, tolman%asylum.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
|> 
|> You are put in a room by evil Nazi's.  There is a button in front of you.  If
|> you press the button, you will die.  However, there are 100 other people in
|> similar rooms.  If no one pushes their button before 12:00, everyone dies.  You
|> have no way to know if the other people have pushed theirs or not. 
|> It is now 11:50. What do you do?
|> 
|> (For more similar problems, read William Poundstone's "The Prisoner's Dilemna")
|> 
|> There is a rational solution.
|> 

It depends who's in the other rooms.  If they are strangers, then no, I don't
think I push the button.  If my wife or child was in one, I definitely push
it, and pray that they had enough faith in me to expect me to have done so.

What a depressing puzzle.

-------------------------------------------------------------------------------
Ronnie B. Kon           |  "If you vote Republican, does that make you
ronnie@cisco.com        |   an accomplice to their crimes?"
(415) 688-4592          |                  -- Bongo, Life in Hell
-------------------------------------------------------------------------------
From: rberesfo@cix.compulink.co.uk (Ross Beresford)
Newsgroups: rec.puzzles
Subject: Re: Crossword Puzzle Software?
Date: 6 Apr 92 18:13:56 GMT
Reply-To: Ross Beresford <rberesfo@cix.compulink.co.uk>
Organization: Gated to News by demon.co.uk

I don't know whether this is precisely what you're looking for, but I am
just completing the development of a crossword-related package for the
PC. It is intended to be helpful in compiling crossword puzzles, as well
as in solving them.

The program searches for entries in a dictionary that match a particular
pattern. It is analogous to programs like GREP, but tailored specifically
for use with crosswords. It has the following features:

    Template matching, anagram matching, and combined template and
    anagram matching. Punctuation and letter case are ignored in matching.

    Elements of the template/anagram pattern can be a single letter, a choice
    of letters, a match-all, a wildcard, or a nested pattern. I've tried to
    design the pattern syntax so it can handle all eventualities!

    I've also put a lot of effort into performance, particularly for the most
    commonly used pattern types.

    There's a simple text mode user interface, and the program can also be
    run from the command line.

The program is called The Electronic Alveary (or TEA for short); it will be
freeware and version 1.00 will be released in a couple of weeks time. I'll
post a message here as soon as it's available from SIMTEL.

Ross Beresford
rberesfo@cix.compulink.co.uk (you're supposed to use this one)
rberesfo%cix@uknet.ac.uk (if the above fails)
Newsgroups: rec.puzzles
From: bmesmer@occs.cs.oberlin.edu (Carl Muckenhoupt)
Subject: Re: Missing subject header
Organization: Oberlin College Computer Science
Date: Thu, 9 Apr 1992 23:04:32 GMT

In article <6052425@MVB.SAIC.COM> Demoel@Fwva.Saic.Com (Ed de Moel) writes:
   news@watson.ibm.com (NNTP News Poster) writes:
   >
   >Prove or disprove that the number of primes in the sequence
   >1 , 11 , 111 ,  1111 , 11111 , ...
   >is finite. 

   Since 111 is divisable by 3 and 1111 is divisable by 11 that shouldn't
   be too hard...
Why? If we restated the problem as "Prove or disprove that the number of
primes in the sequence 1, 2, 3, 4, 5, ... is finite" you could just
as easily say "Well, 8, 9, and 10 are all composite, so it shouldn't be
too hard..." Proof by carefully chosen examples is a dangerous thing.
--

Carl Muckenhoupt        bmesmer@occs.cs.oberlin.edu
"As a lyricist you become sensitive to certain things. You become aware
that there aren't very many words that rhyme with 'love'. There are a lot
more words that rhyme with 'hate' than with 'love'. You've got 'above'
and 'dove' and 'glove' and 'shove' and that's about it."
                        - Peter Shickele
From: fca00@chennai.ccc.amdahl.com (A. Francis Chandran)
Newsgroups: rec.puzzles
Subject: Re: A question of time
Date: 16 Apr 92 22:47:04 GMT
Organization: Amdahl Corporation, Sunnyvale CA


Well here's a wierd possibility ;-)




The clock at the top of the stair is really an anti-clock that
goes anti-clockwise and whose digits are arranged anti-clockwise.

Say the time is 10:30 then the professor thinks that the time
is 1:30. 

He goes downstairs after 5 minutes and sees the time as 1:35
which is 3 hours and five minutes ahead of what he saw upstairs.


                                                 francis.
From: warwick@cs.uq.oz.au (Warwick Allison)
Newsgroups: rec.puzzles
Subject: Re: Brick Problem (crude ASCII)
Keywords: brick, graphical, puzzle
Date: 7 Apr 92 01:17:37 GMT
Reply-To: warwick@cs.uq.oz.au

In <orange.702580580@cunews> orange@alfred.carleton.ca (Paul Vallee) writes:

>     a    b     c
>  |----|-----|----|
>d |  e |  f  |  g | h
>  |----|--|--|----|
>i |   j   |    k  | l
>  |----|--|--|----|
>m |  n |  o  |  p | q
>  |----|-----|----|
>     r    s     t

fjnopkfejokgfblkjimnrtpqhgcaedso


Warwick
--
  _-_|\      warwick@cs.uq.oz.au            / pinots pintos /
 /     * <-- Computer Science Department,  / piston pitons /
 \_.-._/     University of Queensland,    / points potins /
      v      Brisbane, Australia.        / ------ ------ /
From: abbey1@husc9.harvard.edu (Joshua Abbey)
Newsgroups: rec.puzzles
Subject: word squares...
Date: 8 Apr 92 14:23:26 GMT
Article-I.D.: husc3.1992Apr8.102327.10838
Organization: Harvard University Science Center


	Hi there...

	If anyone can help me out with these, I'd appreciate it...

	I have two 3x3 letter matrices that need to be put together to form
	words...you all know the routine...


		Number 1:   A   I   N
			    O   T  (E)
			    B   E   N

		
		Number 2:   T  (A)  G
			    E   E   T
			    O   L   E

		The two letters in parentheses can not be moved...
		(and yes, these are two separate puzzles, don't try to
		combine the two matrices, it won't work...)

	The idea is that they should form actual words, across and down...

		Any solutions, drop me some e-mail...
			(or post it, if you want...i'm not fussy...)

					-Josh
					abbey1@husc.harvard.edu

-- 
==)--------------      abbey1@husc9.harvard.edu   (Josh Abbey)
                "Ashes and diamonds, foe and friend,
                 We were all equal in the end."  - Pink Floyd 
	                 		     	--------------(==
From: jtchern@ocf.berkeley.edu (Joseph Hernandez)
Newsgroups: rec.puzzles
Subject: Re: A simple sequence of numbers ...
Summary: FOR ONCE AND FOR ALL
Date: 9 Apr 92 17:44:15 GMT
Article-I.D.: agate.s1vpfINN1v6
Organization: U.C. Berkeley Open Computing Facility
NNTP-Posting-Host: hailstorm.berkeley.edu

In article <mblum.702838289@chaos.cs.brandeis.edu> mblum@chaos.cs.brandeis.edu (Cerebus) writes:
>meng@idt.unit.no (Tore Mengshoel) writes:
>>In article <1992Apr7.220710.27512@Cadence.COM>, hemmady@Cadence.COM (Shankar Hemmady) writes:
>>|> Continue this sequence:
>>|> 	3, 3, 5, 4, 4, 3, 5, 5, 4, 3, ...
>>e.g. 3, 5, 4, 4, 3, 3, 5, 5, 4, 3, .... 
>What do you mean, "e.g. . . ." ??
>The answer is:
>6, 6, 8, 8, 7, 7, 9, 9, 6, . . .
>Isn't it??

NO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

It has already been established that the sequence goes like this:

		6, 6, 8, 8, 7, 7, 9, 8, 6, ...


NOW CAN WE GO ON TO SOMETHING ELSE? THANKS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Just my 8% of a quarter's worth--

Joe

-- 
-------------------------------------------------------------------------------
*****  *****  *****   Joseph TeleCommunications Enterprises World Headquarters 
    *    *    *                 One Telecommunications Complex Center         
    *    *    *      University of California, Berkeley, California, USA  94720
*****    *    *****         (510) 841-1127   <<jtchern@ocf.Berkeley.EDU>>      
-------------------------------------------------------------------------------
From: ethc_ltd@uhura.cc.rochester.edu (Ethan Corona)
Newsgroups: rec.puzzles
Subject: Re: Math (?) puzzle
Date: 16 Apr 92 01:38:22 GMT
Organization: University of Rochester - Rochester, New York

Why must we be tortured with this puzzle all the time?  I'd like to know.

Ethan Corona
Newsgroups: rec.puzzles
From: vivek@ecst.csuchico.edu (Vivek R. Prabhu)
Subject: Re: A (Not So Obviously) simple number puzzle
Organization: California State University, Chico
Date: Wed, 08 Apr 1992 00:22:11 GMT

In article <1992Apr7.202400.26466@Cadence.COM> hemmady@Cadence.COM (Shankar Hemmady) writes:
>
>The puzzle goes as follows:
>
>"My child's age raised to itself gives my age.
>If my wife is younger than me by my child's age
>and if we are LIVE humans, then what are our ages?"
>
>a small clarification:  if my child's age is x,
>then my age is x**x (x to the power of x)
>And my wife is younger than me by x years.
>
>
>- Shankar Hemmady
>hemmady@cadence.com
>

If we let the wife's age = y yrs.
Then the wife's age i.e. y = (x**x - x) yrs.

Hence we have an equation y = x(x**(x-1) - 1) yrs.

This is representative of the equation (x**2 - 1)

Hence (x - 1) = 2

 .
. . x = 3 yrs.

Taking this as the age of the child, y = 3(3**(3-1) - 1)

Hence y = 24 yrs.

Hence the husband's age is 24 + 3 = 27 yrs.

Boy!!! A rather young age to have a 3 yrs old child.

Regards,
Vivek.
 
-- 
vivek@cscihp.ecst.csuchico.edu (Vivek R. Prabhu),
Research Assistant, Department Of Computer Science, 
School Of Engineering, Computer Science And Technology,
California State University, Chico, California.
Newsgroups: rec.puzzles
From: benham@anion.serum.kodak.com (Todd Benham (x37354))
Subject: wierd birthdays
Organization: Clinical Products Division, Eastman Kodak Company
Date: Fri, 10 Apr 1992 15:12:36 GMT

I am 15 minutes older than my twin brother (actually, I dont have a brother
but let's pretend) but according to our birth certificate's time of birth my
birthday comes one day after my twin brother's and
in different years.  How did this happen (and when are our birthdays)?


If you dont like this puzzle (DILLIGAF), flame my brother, not me.

Todd
Newsgroups: rec.puzzles
From: csmith@onetouch.COM (Craig Smith)
Subject: Re: How do you number a dice?
Organization: MCS/OneTouch, Inc.
Date: Wed, 08 Apr 92 23:52:31 GMT

hemmady@Cadence.COM (Shankar Hemmady) writes:



>This is a puzzle on combinatorics.

>You are given several unnumbered cubical Dices (sp?).  Your objective is to
>number the faces of each dice from 1 to 6.

>If the dices look identical then how many unique dices can you construct?

>If you give me two dices with presumably different numberings, then I must
>not be able to find a one-to-one correspondence (or an equivalence relation)
>between the faces numbered on the two dices.



>A related but simpler problem is:

>	Instead of a cubical dice, if I were to give you several tetarhedral
>(regular pyramids) dices, then how many unique numbering schemes can I assign
>to the four faces of such dices?  
>(Incidentally, the soultion to this problem is used in stereochemistry)


If one assumes that the arrangement of the numbers on the face of the
dice is irrelevant, then the answer is:
for the cube: 30
for the tetrahedron: 2

HOWEVER:

If the way the numbers are written on the side matter, you could have
any number of possibilites in both cases, depending on the way the
numbers are put on the sides.
-- 
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: Craig Smith                                         csmith@onetouch.COM ::
:: Columbus, OH                                  ...!uunet!onetouch!csmith ::
::::::::::::::: "Life is a fulltime occupation."--Gene Kalman :::::::::::::::
Newsgroups: rec.puzzles
From: tolman%asylum.utah.edu@cs.utah.edu (Kenneth Tolman)
Subject: Re: Monday morning brain energizer April 6th
Date: 6 Apr 92 19:39:22 MDT
Keywords: A weekly series in the making?
Organization: University of Utah CS Dept
Distribution: na

In article <1992Apr6.160440.28526@clipper.ingr.com> kontos@clipper.ingr.com (Thorne Kontos) writes:
>
>        The Deception Test
>
>A,B,C,D, and E have just finished taking a test for Deception.
>In their usual chatty way, they are making remarks about their
>places. The remarks of the first and second place finishers
>are false, while the others are true. Lets listen in now.
>
>A: "D was third"
>B: "E was not first"
>C: "I was not last"
>D: "C was lower than B"
>E: "B was second"
>
>What were their places?

(SPOILER)
























































C: saying I was not last, he could not be in the first two.  Therefore
   in position 3 or 4

B: saying that E was not first, with E saying B was second, B can not be
   among the first two.  Therefore in position 3,4 or 5.

A: If D were 3rd now, then C,B and D would be the last three, requiring that
    A was truthful.  D can not be 3rd, so A is in position 1 or 2.

E: Saying that B was second now must be a lie, so E is 1 or 2

D: Must be truthful now, so C is lower than B

We are left with:

1:A
2:E
3:B
4:C  
5:D
From: tolman%asylum.utah.edu@cs.utah.edu (Kenneth Tolman)
Newsgroups: rec.puzzles
Subject: Calendar Cubes
Date: 8 Apr 92 21:15:15 GMT
Article-I.D.: hellgate.1992Apr8.151516.9320
Organization: University of Utah CS Dept


 Some companies manufacture calendar cubes.  These are cubes that
sit in a holder to show the day.  The front faces of the cubes 
are the faces that are seen.


Puzzle 1: (Easier)

How many cubes do you need, and what numbers are on each?

---------------------------------------------------------------
Puzzle 2: (Harder)

How should these numbers be arranged on the cubes so as to
minimize movements?  After each day, the day must be changed
by some combination of turning, switching positions of these
cubes.  Define a movement to be either - switching positions
so as to keep the same faces forward, or a quarter turn of a 
single cube about any axis.

How do we arrange the numbers so as to minimize movements to
rotate through all the days of the year?


---------------------------------------------------------------
Puzzle 3: (Easier)

If you want to show the MONTH, DAY and YEAR, how many cubes
do you need and what numbers should appear on each?
From: wenner@einstein.eds.com (Rich Wenner)
Newsgroups: rec.puzzles
Subject: Re: Math Problem
Date: 8 Apr 92 16:29:47 GMT
Organization: As little as possible.

In article <9204071337.AA25632@ucbvax.Berkeley.EDU> BANDA@AUSVM1.VNET.IBM.COM ("Venu Prabhakar Banda") writes:
>Problem: Show that (4444 ** 4444) is divisible by 9

I assume that '**' is supposed to represent exponentiation,
a la FORTRAN?

4444 factors as 2 x 2 x 11 x 101, and you can exponentiate that
to hell and back and it will never be divisible by 9.

Is this some kind of sleazy trick based on not telling us what
the number base is, or some such thing?

Let's add another relatively trivial problem:  assuming no sleazy
tricks, what is (4444 ** 4444) mod 9?  This should be cake without
resorting to either serious math or computer programs, especially
if you've been reading the group lately.


I get 7.
-- 
Rich Wenner                   --(o)--(o)--     I really hate this damn machine.
EDS/TSD/DSD/FSD/CSDPS         (    /\    )     I wish that they would sell it.
                                \ mmmm /       It never does quite what I mean,
#include <stddisclaimer.h>       \____/        But only what I tell it.
Newsgroups: rec.puzzles
From: mathur@nas.nasa.gov (Sanjay R. Mathur)
Subject: Tough Cryptics
Organization: NASA Ames Research Center, Moffett Field CA
Date: 16 Apr 92 22:16:37


    This month's Atlantic magazine has an advertisement for a 
publication called "Tough Cryptics" and I was wondering if anyone here
has had a chance to go through it? Is it worth the cost ? ( The subscription
rate  is somewhat expensive, @ $27 for six issues) Do they have standard 
British newspaper style crosswords? I enjoy the Atlantic's Puzzler as
well as the cryptic in Harper's but I don't think trying more than two
of their kind per month would be good for one's sanity -:)

   On a related note, I recall someone on this newsgroup asking
recently for an introduction to cryptic clue solving. The Atlantic offers
just such a thing; in return for a self-addressed, stamped envelope to
The Atlantic Puzzler, 745 Boylston Street, Mass. 02116. I have not
seen it but I have found their clues at least to be in the conventional
cryptic style and so I assume it will be a good starting point for
beginners. 

--

Sanjay Mathur                                              mathur@nas.nasa.gov
MS 258-2                                                    (415)604-3958 (W)
NASA Ames Research Center                                   (415)969-0209 (H)
Moffett Field, CA 94035


Newsgroups: rec.puzzles
From: uge@athena.mit.edu (Eugene A Beidl)
Subject: Re: trvia (alternative spoiler)
Organization: Massachusetts Institute of Technology
Date: Mon, 13 Apr 1992 23:55:20 GMT

In article <djoyce.703038701@black.clarku.edu>, djoyce@black.clarku.edu (Dave Joyce) writes:
|> In <703024532.1@nwark.FidoNet> Marc.Costaldi@f2050.n391.z1.FidoNet.Org (Marc Costaldi) writes:
|> 
|> >Got a problem need some help with:
|> >
|> >An architect was asked to design a square house with the window in each
|> >wall 
|> >having a southern exposure. How was this done?
|> 
|> Alternative spoiler:
|> 
|> 
|> 
|> 
|> 
|> Let's see now.  Suppose there was only one wall having a southern exposure.
|> Then the architect would only have to put a window in that wall.  So, he/she
|> would design a square house with no windows on the north, east, or west
|> walls, but one window on the south wall.  That explains why the problem
|> refers to "the" window: there only is one window.

I disagree.  The problem refers to "the window" in EACH wall.  Thus each wall
has only one window.  I believe the other answer is the only correct one.


--Eugene A. Beidl
--uge@athena.mit.edu
Newsgroups: rec.puzzles,soc.culture.indian
From: russotto@eng.umd.edu (Matthew T. Russotto)
Subject: Re: A simple number puzzle
Date: Thu, 09 Apr 92 15:00:49 GMT
Organization: College of Engineering, University of Maryland, College Park

In article <1992Apr8.042047.20639@wuecl.wustl.edu> bs@wucs1.wustl.edu (Bala SWAMINATHAN) writes:
>
>Did the original article imply any where that LIVE humans can have only
>integral ages? OR did the very use of the word *age* imply an integer?
>
>If the age can be counted in months (say 26 years and 5 months etc.)
>then there are more solutions.

Everyone I know, when asked their age, will either give a
non-numerical answer ("What business is it of yours?" or some variant
:-) ) or will state their age in years.



-- 
Matthew T. Russotto	russotto@eng.umd.edu	russotto@wam.umd.edu
Some news readers expect "Disclaimer:" here.
Just say NO to police searches and seizures.  Make them use force.
(not responsible for bodily harm resulting from following above advice)
Newsgroups: rec.puzzles
From: wyg9633@uxa.cso.uiuc.edu (Wooed Sun)
Subject: Re: Math Problem (Solution?)
Organization: University of Illinois at Urbana
Date: Wed, 8 Apr 1992 21:22:49 GMT

vec@sunshine.Kodak.COM (Vance Cochrane) writes:

>In article <1992Apr8.160455.24623@ux1.cso.uiuc.edu>, wyg9633@uxa.cso.uiuc.edu (Wooed Sun) writes:
>.
>.
>>The base is not necessarily refined to 8 for the number to be divisible
>>by 9. Let's say the base is x and assume x > 9. Then,
>>4444 = 4*(x**3+x**2+x+1) = 4*(x**4-1)/(x-1)
>>If you check with your calculator, you can see 4444 is divisible by 3

>what base are you discussing here? try to be a little more precise since it
>is obviously not base 10!

OK, here goes a clarification:

   4444   = [4*(x**3+x**2+x+1)]   = [4*(x**4-1)/(x-1)]
       (x)                    (10)                   (10)

>>(therefore, 4444**2 is divisible by 9, and so is 4444**4444) when
>>x = 11,12,14,17,...

Actually, x = 11,14,17,...

>-- 
>Vance Cochrane			Email: vec@Kodak.COM
>Research Laboratories		Voice: (716) 722-3709
>Eastman Kodak Company		Fax:   (716) 722-0160
>Rochester, NY 14650-1816
>USA
-- 
email : w-yang@uiuc.edu (wyg9633@uxa.cso.uiuc.edu)
Newsgroups: rec.puzzles
From: jwd2@arcadien.owlnet.rice.edu (Joseph William DeVincentis)
Subject: Re: The negative Harmonic series
Organization: Rice University
Date: Sun, 12 Apr 1992 19:32:07 GMT

In article <1992Apr11.020521.16082@Cadence.COM>, hemmady@Cadence.COM (Shankar Hemmady) writes:
> 
> 
> The series that David G. Caraballo refers to is the well-known
> harmonic series with alternately positive and negative terms.
> 
> It has puzzled mathematicians for a long time.
> 
> It satisfies the rules of convergence, but not those of absolute
> convergence.
> 
> So here is a puzzle on that subject.  (David please do not answer this
> one)
> 
> 	1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 ....
> 
>    =	1 - (2/2-1/2) + 1/3 - (2/4-1/4) + 1/5 - (2/6-1/6) + 1/7 - (2/8-1/8) ....
> 
> 	expanding the brackets and re-grouping ...
> 
>    = 	(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...)
> 	- (2/2 + 2/4 + 2/6 + 2/8 + ...)
> 
> 	factorizing ther 2/2, 2/4, etc ...
> 
>    = 	(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...)
>      	- (1 + 1/2 + 1/3 + 1/4 + ...)
> 
>    =    ZERO!!
> 
> 
> What went wrong.  I know that there is the Absolute Convergence theorem which
> can find the error in the above statements.  But can you give a 
> simple lay-man explanantion of the fallacy?

I'd explain it by saying that in the final step before you get zero, you
have infinity minus infinity, which is not necessarily zero, but rather
an undefined quantity, similar to 0/0.

-- 
\  ,------------------------------------------------------------.  /
 \/     Joseph  De Vincentis            jwd2@owlnet.rice.edu     \/
 /\           Help! My .sig just flipped umop-ap!sdn !           /\
/  `------------------------------------------------------------'  \