HDTenEightyP/NetNews huggingface dataset:
From: dmccart@gomez.intel.com (D. J. McCarthy ~)
Newsgroups: rec.puzzles
Subject: Re: Math puzzle
Date: 27 Jul 92 20:57:24 GMT
Distribution: rec.puzzles
Organization: Architecture & Software Technology Group, Intel, Santa Clara CA
X-Newsreader: Tin 1.1 PL4

amlogan@bird.Princeton.EDU (Adam M. Logan) writes:

> More generally, (1/a + 1/(a + b) + 1/(a + nb)) is never an integer.


	...for nonzero b.  If b=0 then (1/a + 1/(a + b) + 1/(a + nb)) is
an integer for four values of a.  (Which four values is left as an
exercise for the reader's five-year-old.  :-)  )



--
D. J. McCarthy <dmccart@gomez.intel.com>
"Klingons who have never left the [home] planet are called Static Klingons.
 A good dryer sheet or fabric softener is known to get rid of them."
                            --Terry J. Wood, rec.arts.startrek.current, 7/18/92
From: dseal@armltd.co.uk (David Seal)
Newsgroups: rec.puzzles
Subject: Re: Infinite series and bogus math proofs
Date: 30 Jul 92 14:59:05 GMT
Distribution: rec
Organization: A.R.M. Ltd, Swaffham Bulbeck, Cambs, UK

In article <1992Jul29.162316.14848@CSD-NewsHost.Stanford.EDU>
amorgan@Xenon.Stanford.EDU (Crunchy Frog) writes:

>To combine two popular (!) threads in this newsgroup I submit the following
>proof that infinity = -1
>
>S = 1 + 2 + 4 + 8 + ........       
>
>2S = 2 + 4 + 8 + 16 + ........
>
>S-2S = 1
>
>S = -1

Ever wondered why most computers represent the integer -1 as a word with all
bits set? Under the normal interpretation of binary numbers, this word
represents:

  1 + 2 + 4 + 8 + ... + 2^(word length - 1)

This is the best representation the computer can manage of S = 1 + 2 + 4 + 8
+ ..., and according to the above, S = -1.

:-) :-) :-)

If you like these, you might also consider:

  T  = 1 + 4 + 16 + 64 + ...

  2T = 2 + 8 + 32 + 128 + ...

  3T = 1+2+4+8+16+32+64+128+...
     = S
     = -1

  T  = -1/3

:-) :-) :-)
 
There is in fact some serious mathematics behind all this. If you're
interested, look up "p-adic numbers": the above are effectively valid proofs
about 2-adic numbers. Where they go wrong is if you think that they are
proofs about ordinary numbers, and *especially* if you think that infinity
is an ordinary number...

David Seal
dseal@armltd.co.uk

All opinions are mine only...
Newsgroups: rec.puzzles
From: Ulf.Lunde@kvatro.no (Ulf Lunde)
Subject: 3 puzzles - Sorry, my mistake!
Organization: Kvatro A/S, Trondheim, Norway
	<5084@transfer.stratus.com> <1992Jul26.134105.13917@ens.fr>
	<1992Jul27.145517.7957@cbfsb.cb.att.com>
	<ULF.LUNDE.92Jul31210354@hugin.kvatro.no>
Date: 31 Jul 92 22:18:36

 On 31 Jul 92 21:03:54, Ulf.Lunde@kvatro.no (I) said:
>>>>	What is the first number, when spelled out (e.g. O-N-E) is
>>>>in alphabetical order?
>>>>40  (F-O-R-T-Y)
> 
> Forty??   How about F-I-V-E ?   Or F-I-F-T-E-E-N?   Or E-I-G-H-T ?

OK, please don't flame!  I didn't read the question thoroughly enough:  
I read the "is" as a comma!  :-)


>>>Is there a language in which 1 is so ?
>> 
>>   Eins in German.
>> 
>> Mazen Mokhtar
> 
> Oh yeah?  Just how do you spell e.g. 397 in German, then?  Or 8?

Obviously not with the letters in alphabetical order, admittedly 
--- see above.  Thanks to Mazen for explaining it.


> I Norwegian I think 18 must be the first one: A-T-T-E-N.

But 1 (E-N) looks so trivial...  :-b


> In Esperanto it is 10 (D-E-K) if you don't count B-I-L-I-O-N-O (1E12).

No, in Esperanto it is 2 (D-U).  That too looks trivial...


> How many numbers (in how many languages) tell you how many letters
> they are made up of?  E.g. F-O-U-R in English.

T-O in Norwegian, D-U in Esperanto.

Ulf Lunde
From: hoey@AIC.NRL.Navy.Mil (Dan Hoey)
Newsgroups: rec.puzzles
Subject: Re: Plane Division
Date: 23 Jul 92 11:21:32 GMT
Organization: Naval Research Laboratory, Washington, DC

s290@brems.ii.uib.no ( Per Ivar Roervik) writes:

>  In a closed and bounded area of any 2-D shape, 1 million random
>  points are chosen.
>  Is it always possible to bisect the area AND the points into two
>  equal parts with a straight line?

Several respondents have offered counterexamples, but they all involve
lots of collinear points and the inability to avoid some of them.

Suppose we allow a line through some points to `divide' those points
in whatever way makes the overall division most even.  So if a line
goes through 500,000 points, it will perforce divide the set of one
million in half.

I am pretty sure the answer in this setting is yes.  It's essentially
a kind of ham sandwich theorem.  The ham sandwich theorem says that no
matter how sloppily you make a ham sandwich, it's always possible with
one knife cut to bisect both the ham and the bread.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From: mlevis@ringer.cs.utsa.edu (Mike Levis)
Newsgroups: rec.puzzles
Subject: Re: Ten equals 9.9999....
Date: 30 Jul 92 03:53:58 GMT
Organization: University of Texas at San Antonio

In article <1992Jul29.174207.5974@ecst.csuchico.edu> jchamber@ecst.csuchico.edu (John William Chamberlain) writes:
>       if 1/3 + 1/3 + 1/3 = 1
>       
>       and .33... + .33... + .33... = .99999
>  
>       since .33.... = 1/3
>
>       1/3 + 1/3 + 1/3 must equal .99999..
>
>       therefore
>        
>       .99999... = 1/3 + 1/3 + 1/3 = 1 = .33... + .33... + .33...
>
>       and thus
>  
>       .999... = 1

This seems to be the best proof so far, and it is so simple.  I
have been neutral on this so far, but that ``.3333... + .3333...
+ .3333... = .9999...'' part convinced me that .9999... = 1.0

Just my 1.9999... cents.

>  John

-- 
========= Mike Levis   mlevis@ringer.cs.utsa.edu =========  .--.
ftp syrinx.umd.edu (128.8.2.114) for Rush lyrics and info. (OS/2)
::::: Get the Rush FAQL in directory /rush/special. ::::::  ~--~
Newsgroups: rec.puzzles
From: amorgan@Xenon.Stanford.EDU (Crunchy Frog)
Subject: Re: Alien colonization
Organization: Computer Science Department, Stanford University.
Date: Wed, 22 Jul 1992 22:11:02 GMT

In article <1992Jul22.153416.3564@hellgate.utah.edu> 
  tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:

>  The alien Ddiv lives in a universe which is three dimensional, and has
>the planets evenly spaced throughout as in a 3D grid.
>Ddiv is planning to have children- each child will require one full planet
>to live on.  Ddiv is also immortal, and wishes to keep living forever.
>He wishes to be fair to his children- so he will only have as many 
>children as each of his children can have.  He would not have any children
>if any single one of his offspring could not have any children, so
>his grandkids, great grandkids, and so on must be able to have as many
>children as he has. 
>
>
>  How many children can Ddiv have? (Ddiv has children by dividing)

Is the trick with this one that if Ddiv has >1 child after an infinite
amount of time there will be an uncountably infinite number of Ddiv'lets
and thus not enough planets for them to live on?

I say one child (but I would be surprised if I am right).

C Frog
Newsgroups: rec.puzzles
From: jimhsu@athena.mit.edu (Jimmy M.D. Hsu)
Subject: Just wondering... Clarification
Organization: Massachusetts Institute of Technology
Date: Wed, 29 Jul 1992 21:27:53 GMT

I need to mention that in this
case, the three face cards must
appear for each player, not for
the whole group.

Sorry if you're confused.
(or maybe...  :-)


--
James AMDEE   MIT

For a price I'd do about anything.

Except pull the trigger.





For that I need a pretty good cause...


			-Queensryche.
Newsgroups: rec.puzzles
From: deghare@daisy.waterloo.edu (Dave Hare)
Subject: Re: Proof of -1 = 1
Organization: University of Waterloo
Date: Fri, 24 Jul 1992 18:36:44 GMT

In article <1992Jul24.155639.11086@eng.umd.edu> rabino@eng.umd.edu (Michael Rabinovitz) writes:
>In article <chan.711960216@s1.elec.uq.oz.au> chan@elec.uq.oz.au (Chuen Wen Chan) writes:
>>
>>>	sqrt(-1) = sqrt(-1)
>>
>>>	sqrt( -1 / 1 ) = sqrt( 1 / -1 )
>>
>>>	sqrt(-1)    sqrt(1)
>>>	-------- = ---------
>>>	sqrt(1)     sqrt(-1)
>>
>>>	now cross multiply and we get:
>>
>>>	-1 = 1  QED
>>
>>Regarding the above problem, I DO know that the third line is 
>>wrong (ie: i != 1/i).  The point is the derivation up to that
>>point looks legal.
>>
>	IMHO, the reason line 3 is wrong is the expansion of line 2
>assumes 
>		sqrt(a/b) = sqrt(a) / sqrt(b) 
>rather than
>		sqrt(a/b) = sqrt(a) * sqrt(1/b).
>
>	Using this form to evaluate line 2 makes the "paradox" go away.
>
>					Mike Rabinovitz
>					rabino@eng.umd.edu

No, the problem really depends on what is meant by "sqrt".  

If you decide ahead of time what the value of  sqrt(x)  will be for every
x  in the complex plane (e.g., if you pick the principal branch), then
the transformation  sqrt(a*b) --> sqrt(a)*sqrt(b)  is not valid for all
points  a  and  b .  (Hence, neither is the transformation  
sqrt(a/b) --> sqrt(a) * sqrt(1/b)  valid in general).

However, if you let  sqrt(x)  stand for "any (or all) possible numbers whose
squares give  x ", then there is no error until the last step above, which
should, in this case, read  {-1,1} = {-1,1} .

Dave Hare
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Gummy Drop-bears <SPOILER>
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Thu, 30 Jul 1992 15:48:00 GMT

In article <Jul.29.03.30.41.1992.18050@remus.rutgers.edu> clong@remus.rutgers.edu (Chris Long) writes:
>
>Real gummy drop-bears mass 10 grams, while imitation gummy
>drop-bears mass 9 grams.  Spike has 7 cartons of gummy drop-bears,
>4 of which contain real gummy drop-bears, the others imitation.
>Using a scale only once and the minimum number of drop-bears, how
>can Spike determine which cartons contain real gummy drop-bears?

<SPOILER> ahead...




Spike used 51 gummi-drop bears:  from the 7 boxes he took respectively
0, 1, 2, 4, 7, 13, and 24 bears.

The notion is that each box of imitation bears will subtract its
number of bears from the total "ideal" weight of 510 grams (1 gram of
missing weight per bear), so Spike weighs the bears, subtracts the
result from 510 to obtain a number N, and finds the unique combination
of 3 numbers from the above list (since there are 3 "imitation" boxes)
that sum to N.

The trick is for the sums of all triples selected from the set S of
numbers of bears to be unique.  To accomplish this, I put numbers into
S one at a time in ascending order, starting with the obvious choice,
0.  (Why is this obvious?  If I'd started with k > 0, then I could
have improved on the resulting solution by subtracting k from each
number) Each new number obviously had to be greater than any previous,
because otherwise sums are not unique, but also the sums it made when
paired with any previous number had to be distinct from all previous
pairs (otherwise when this pair is combined with a third number you
can't distinguish it from the other pair)--except for the last box,
where we can ignore this point.  And most obviously all the new
triples had to be distinct from any old triples; it was easy to find
what the new triples were by adding the newest number to each old sum
of pairs.

Now, in case you're curious, the possible weight deficits and their
unique decompositions are:

3	= 0 + 1 + 2
5	= 0 + 1 + 4
6	= 0 + 2 + 4
7	= 1 + 2 + 4
8	= 0 + 1 + 7
9	= 0 + 2 + 7
10	= 1 + 2 + 7
11	= 0 + 4 + 7
12	= 1 + 4 + 7
13	= 2 + 4 + 7
14	= 0 + 1 + 13
15	= 0 + 2 + 13
16	= 1 + 2 + 13
17	= 0 + 4 + 13
18	= 1 + 4 + 13
19	= 2 + 4 + 13
20	= 0 + 7 + 13
21	= 1 + 7 + 13
22	= 2 + 7 + 13
24	= 4 + 7 + 13
25	= 0 + 1 + 24
26	= 0 + 2 + 24
27	= 1 + 2 + 24
28	= 0 + 4 + 24
29	= 1 + 4 + 24
30	= 2 + 4 + 24
31	= 0 + 7 + 24
32	= 1 + 7 + 24
33	= 2 + 7 + 24
35	= 4 + 7 + 24
37	= 0 + 13 + 24
38	= 1 + 13 + 24
39	= 2 + 13 + 24
41	= 4 + 13 + 24
44	= 7 + 13 + 24

Note that there had to be (7 choose 3) distinct values; they end up
ranging from 3 to 44 inclusive with 7 numbers missing:  4, 23, 34, 36,
40, 42, and 43.

-- David Karr (karr@cs.cornell.edu)
From: fxsdb@acad3.alaska.edu (Scott D. Berry)
Newsgroups: rec.puzzles
Subject: Re: Easy prob. question
Date: 25 Jul 92 03:53:32 GMT
Organization: University of Alaska Fairbanks

In article <1992Jul25.022553.5173@beaver.cs.washington.edu>, phaedrus@cs.washington.edu (Mark Phaedrus) writes:
> In article <1992Jul24.160443.1@acad3.alaska.edu> fxsdb@acad3.alaska.edu (Scott D. Berry) writes:
>>In article <1992Jul24.225745.19641@EE.Stanford.EDU>, mack@isl.Stanford.EDU (mack) writes:
>>> This looks to be an easy one:
>>> Q: Two cards are picked at random from a complete deck of cards. What
>>>    is the probability that they are of the same suit?
>>In fact, this is an easy one.  The answer depends on a couple things, tho.
>>What kind of deck (a pinochle deck)?  Assume a poker deck, complete.  The
>>chances are (52/54) * (13/54) + (2/54)^2, if two jokers are considered to 
>>be the same suit.
> 
>      And, like most easy ones, it's very easy to slip up and provide a wrong
> answer. :)  Using your assumptions (54-card deck, 2 jokers, jokers match each
> other but not any other card in the deck), the odds look to me to be:
> 
>  52 non-jokers   12 other cards in same suit     2 jokers   1 other joker
> (------------- x ---------------------------) + (-------- x --------------)
>    54 cards            53 other cards            54 cards   53 other cards
>    
> or ((26*12)+1)/(27*53), or 313/1431.
>      Your odds don't account for the fact that the same card can't be chosen
> as both the first and second card.
oops!  Actually, I misread the question, and was pulling one card from each of
two decks.
   -Scott
Newsgroups: rec.puzzles
From: tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman)
Subject: Re: Plane Division
Date: 22 Jul 92 15:19:43 MDT
Organization: University of Utah, CompSci Dept

Consider all possible lines which divide a shape in half ( a regular one for
now).  All of these lines must cross the center if it, for a circle, for a
square, and so on.

  We now will choose all of the points so that the lie along one of these
dividing lines, and are in one half of the figure.  For instance, the
radius of a circle.  

 In order to break the points in half, we must strike them at some angle
which is different than the line they all lie in.  Yet, if we did that
then our line would no longer divide the area in half.

  It is possible to construct a set of points which can not be broken in
half by a line which also divides the figure into two equal areas.

QED
From: lbaum@atc.boeing.com (Larry Baum)
Newsgroups: rec.puzzles
Subject: Re: 3 puzzles
Date: 29 Jul 92 16:26:25 GMT
Organization: Boeing Computer Services

In article <1992Jul26.233848.20722@Armstrong.EDU> water772@Armstrong.EDU (Kelly
Waters) writes:
>>>>>In article <5071@transfer.stratus.com> markc@eomer.sw.stratus.com (Mark
Christensen) writes:
>>>>>
>>>>>>	What is the first number, when spelled out (e.g. O-N-E) is
>>>>>>in alphabetical order?
>>>>>
>
>
>SPOILER ALERT !!!!!!!!!!!
>
>
>
>
>
>
>	F-O-R-T-Y

Well EIGHT is alphabetically before FORTY, but how about:

ALEPH-NOUGHT

(actually, there is probably a definition of ALEPH-EIGHT)
Newsgroups: rec.puzzles
From: kanad@top.cis.syr.edu (Kanad Chakraborty)
Subject: Re: preposition proposition
Organization: Syracuse University, CIS Dept.
Date: Fri, 24 Jul 92 12:44:59 EDT

In article <2113@tymix.Tymnet.COM> filfeit@rael.Tymnet.COM writes:
>Everybody knows that a preposition is a word you should not
>end a sentence with B-).
>
>Can someone post a sentence that ends with FIVE prepostions?
>
>--f2
>
>----------------------------
>  Time flies like an arrow.
>  Fruit flies like a banana.

These are some words you should not end a sentence with - in, on,
at, upon.    :)


OK, so I cheated ! But who cares, if the sentence is meaningful?

Kanad Chakraborty
From: warwick@cs.uq.oz.au (Warwick Allison)
Newsgroups: rec.puzzles
Subject: Re: Spacewarp wonders
Date: 24 Jul 92 01:14:47 GMT
Reply-To: warwick@cs.uq.oz.au

tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:

>  Your spaceship has been sucked through a spacewarp, and then crashed
>onto a place.  The place is foggy, so one can not see more than a few
>kilometers clearly.  There is always a fairly bright light here.
>Also, you have access to whatever science gear you might like, as it
>was aboard your craft.  There are lots of life forms, all of them huge.

>1) How would you be able to tell if the planet was a sphere, or a nearly flat
>   plane extending in all directions?

I would use the Deep Dig core sampling unit, attached the the Gravity
Wave Detector, all kept cool by the Environment Maintenance Pod to make
a map of the gravitational field in a downward direction, and from that
deduce the topology of my location.  eg. if the gravitational field
zeroed out at distance X, then started to increase, I would assume I was
on a roughly spherical object.  If it ground to a halt, imbedded in a
material that was bonded with the strength of neutronium, I would assume
I was on a Niven Ring, or Dyson Sphere.

>2) How would you be able to tell if you had travelled forwards or backwards
>   through time 5 billion years or more?

I would use the Infinite High Sensitivity Radio Receiver to listen for
and measure the dispersion of TV programs from Earth, then look up the dates
for transmission in the Television Almanac Database, and calculate the
distance I was from Earth, and what they were watching that many light years
ago.

>3) How would you be able to tell if you and your space ship were somehow
>   shrunken in passage?

I would simply use the Atomic Caliper to measure the distance between
the atomic and subatomic particles of some atoms of the surroundings,
if they seemed large, my Atomic Caliper has shrunk (in which case, I'd
probably have to recalibrate the Infinite High Sensitivity Radio Receiver).


But of course, all hacker know that ANYTHING is possible... given the hardware.


--
Warwick
--
  _-_|\      warwick@cs.uq.oz.au            /Disclaimer:
 /     * <-- Computer Science Department,  /  
 \_.-._/     University of Queensland,    /      void (if removed)
      v      Brisbane, Australia.        /
From: harmsen@cs.vu.nl (Harmsen H)
Newsgroups: rec.puzzles
Subject: need an algorithm
Date: 31 Jul 92 09:17:41 GMT

I am looking for an algoritm for determining unused ID's.

I've got a list of objects, each with a unique ID. Objects can be
appended and removed.
Each time a new object will be appended it must have a new ID.

Now I need a simple but relatively quick algorithm to determine
an ID which is at that time not used. Just traversing the numbers 1..n
and for each number scan de list to see if it's used is too slow.

Taking the next ID (1, 2, 3, .... 20, 21, .....10000, 10001 etc.) each 
time is of course very fast but will result in an overflow sometime 
in the future.


Hans-Peter Harmsen		(harmsen@cs.vu.nl)
Free University, Amsterdam


P.S. I don't want to keep a list of used ID's.

--
----------------------------------------------------------------------
| Unbelievable but true: The African continent contains so much sand |
| that, if spread out, it would cover the entire Sahara desert.      |
----------------------------------------------------------------------
Newsgroups: rec.puzzles
From: arromdee@jyusenkyou.cs.jhu.edu (Ken Arromdee)
Subject: Re: Spacewarp wonders 
Organization: Johns Hopkins University CS Dept.
Date: Thu, 23 Jul 1992 21:07:22 GMT

In article <1992Jul23.200727.5290@rice.edu> andsol@is.rice.edu (Andrew J. Solberg) writes:
>|> 3) How would you be able to tell if you and your space ship were somehow
>|>    shrunken in passage?
>Assuming that the laws of physics do not change, the Bohr radius of an
>atom would be immutable.  There are a number of experiments we can devise
>to find the Bohr radius accurately.  We then compare the 'found'
>radius to the known value to find whether we have changed in size.

Assuming that the laws of physics don't change, we can't be shrunken in the
first place.  One would think that to be shrunken means that _everything_
is to be shrunken; this includes Bohr radiuses too....
--
Hi! Ani mutacia shel virus .signature. Ha`atek oti letoch .signature shelcha!

Ken Arromdee (UUCP: ....!jhunix!arromdee; BITNET: arromdee@jhuvm;
     INTERNET: arromdee@jyusenkyou.cs.jhu.edu)
From: jesperse@nas.nasa.gov (Dennis Jespersen)
Newsgroups: rec.puzzles
Subject: Rift in the fabric of the space-time continuum?
Date: 24 Jul 92 17:50:13 GMT
Organization: Computational Fluid Dynamics Branch, NASA Ames Research Center


On Monday morning, July 20, 1992, I happened to glance at my digital
watch and noticed that it indicated the date was 7/19.  The time
(hours and minutes) on the watch seemed correct.  "That's odd", I
thought, "I'm here at work so it's probably not Sunday."  I reset the
date to 7/20.  Later that afternoon I noticed that the date on my
watch was 7/21; the time still seemed correct.  I was momentarily
disoriented, not knowing for sure what day it was.  "Maybe there's
been a rift in the fabric of the space-time continuum, and it's really
Tuesday now."  I soon convinced myself that it was still Monday, and I
reset the date again.

Monday evening I glanced at my watch again and suddenly realized what
had happened.  What was the explanation?

Partial spoiler ahead:
On Friday, July 17, three days prior to the indicated events, I had
found it necessary to reset all the functions on my watch.
--
Dennis Jespersen
MS 202A-1
NASA Ames Research Center
Moffett Field, CA 94035-1000
Voice: (415) 604-6742
FAX:   (415) 604-1095
email: jesperse@rft28.nas.nasa.gov
We have met the enemy and he is us. -- Pogo
From: johnson@spectra.com (boyd johnson)
Newsgroups: rec.puzzles
Subject: Re: FAQ???
Keywords: Frequently Asked Questions
Date: 24 Jul 92 21:21:54 GMT
Organization: Spectragraphics Corporation

In article <1992Jul21.155558.7266@CSD-NewsHost.Stanford.EDU> amorgan@Xenon.Stanford.EDU (Crunchy Frog) writes:
>In article <rsakella.711639130@p4.cs.man.ac.uk> 
>  rsakella@p4.cs.man.ac.uk (Rizos Sakellariou) writes:
>>Is there a list of Frequently Asked Questions for rec.puzzles?
>
>That question is on the Frequently Asked Questions list - the answer
>by the way, is NO.
>
>C "The preceeding statement is false" Frog

If you met two natives in the jungle and knew one of them always told the
truth and one of them always lied, what would you conclude about the
availability of an FAQ list if one of them said Frog's above statements?  :-)

-- 
======== Boyd Johnson    nosc!spectra.com!johnson  San Diego, Ca. ========
Newsgroups: rec.puzzles