HDTenEightyP/NetNews huggingface dataset:
From: emru@CAM.ORG (Emru Townsend)
Newsgroups: rec.puzzles
Subject: Re: Alien problem
Date: 31 Jul 92 02:20:44 GMT
Organization: Communications Accessibles Montreal, QC CANADA

In article <Jul.29.14.42.42.1992.2450@remus.rutgers.edu> miklas@remus.rutgers.edu (Keith Miklas) writes:
>Don't even read this one unless you have a few hours on your hands.
>
>Two scientists are walking together discussing how much better an
>editor emacs is than vi.
>
>An alien appears, and says "hi guys, I'm
>thinking of 2 numbers between 3 and 97 (inclusive)."  He then says,
>"The sum of these two numbers is..." and whispers it into the first
>professor's ear.  He then continues, "The product of these two numbers
>is..." and whispers it into the second professor's ear.  The alien then
>vanishes in a cool puff of smoke.
>
>The first professor says to the second, "It is impossible for you to
>know what the two numbers are."
>
>The second arrogantly replies, "Well, now that you've said that, I do
>know what they are."
>
>The first smiles and says, "Well, now I know what they are too."
>
>And they continue their conversation.
>
>
>
>Fellahs, The question is, What are the two numbers?

Pi and something else?

=========================================================================
Emru Townsend             | "Never let a  | /  ___            _    \\  //
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emru@cam.org  BIX: switch | you're in a   |  \  |  | | | | | | |    //\\ 
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=========================================================================
From: joe@csource.oz.au (Joe Slater)
Newsgroups: rec.puzzles
Subject: Tiling a plane
Date: Mon, 20 Jul 1992 09:52:01 +1000

A plane can be tiled with three regular geometric shapes: squares, triangles and
hexagons. It can also be tiled with pairs of shapes: hexagons and triangles, or
octagons and squares.

Are there any other pairs that will tile a plane? What about triplets? Answers
requiring fractal-like series to fill up gaps aren't counted.

jds





 * Origin: What horrors wait for me in this, the Phantom's Opera? (3:632/351)
Newsgroups: rec.puzzles
From: ph@vortex.ama.caltech.edu (Paul Hardy)
Subject: Re: carrying the torch (HINT)
Organization: California Institute of Technology
	<1992Jul25.065442.12748@beaver.cs.washington.edu>
	<PH.92Jul25130436@vortex.ama.caltech.edu>
	<1992Jul26.055648.26658@beaver.cs.washington.edu>
Distribution: rec
Date: Mon, 27 Jul 1992 08:52:16 GMT

In article <1992Jul26.055648.26658@beaver.cs.washington.edu> phaedrus@cs.washington.edu (Mark Phaedrus) writes:

SPOILER FOLLOWS


>  Fine, be that way, see if I care.  :) Well, if I can't get it one way, 
>  I'll get it the other:  Turn two switches (call them #1 and #2) on.  Wait a few
>  minutes, then turn #2 off and race into the other room.  Light #1 will still
>  be on, #2 will be off but warm to the touch, and #3 will be off and at room
>  temperature.  Incandescent bulbs certainly don't cool down that quickly; I've
>  nearly burned my hands on bulbs that have been off for a minute or more.

Yes, this is the answer!  I almost said this after your last post because it
was so close.

				--Paul

--
This is my address:         ph@ama.caltech.edu
This is UUCP:               ...!{decwrl,uunet}!
This is my address on UUCP: ...!{decwrl,uunet}!caltech.edu!ama!ph
Any questions?

"Does Emacs have the Buddha nature?"  --Paul Hardy   "Yow!" --Zippy
From: clong@remus.rutgers.edu (Chris Long)
Newsgroups: rec.puzzles
Subject: Re: Gummy Drop-bears
Date: 30 Jul 92 06:39:59 GMT
Organization: Rutgers Univ., New Brunswick, N.J.

In article <ycpmawg.obrien@netcom.com>, Robert O'Brien writes:

> It is necessary to believe these things are manufactured with far greater
> precision than any *real* confections, but...

These are gummy drop-bears, which are far superior to gummy bears.

{ Method using 63 deleted }

Good try, but little Spike came up with a method that uses fewer than
55 gummy drop-bears.
-- 
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618

"In a study of schoolboys, an educator discovered a correlation between size
 of feet and quality of handwriting.  The boys with the larger feet were,
 on the average, older."        Wallis & Roberts, _The Nature of Statistics_
From: keith@chirality.rsa.com (Keith Vetter)
Newsgroups: rec.puzzles
Subject: Math puzzle
Date: 21 Jul 92 16:00:18 GMT
Article-I.D.: chiralit.KEITH.92Jul21110018
Distribution: rec.puzzles
Organization: RSA Data Security, Inc.


As is well known the sum  1/1 + 1/2 + 1/3 + ... goes to infinity.  The 
question is, does the partial sum  S = 1/1 + 1/2 + 1/3 + ... + 1/n
ever equal an integer value (other than n=1)?

Keith@rsa.com
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Ten equals 9.9999....
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Tue, 28 Jul 1992 15:20:26 GMT

In article <1992Jul27.150032.12515@phillip.edu.au> t9114145@phillip.edu.au writes:
>In article <Bry202.4BF@watdragon.waterloo.edu>, deghare@daisy.waterloo.edu (Dave Hare) writes:
>> In article <1992Jul25.145154.12498@phillip.edu.au> t9114145@phillip.edu.au writes:
>>>In article <1992Jul24.102553.17411@pasteur.Berkeley.EDU>, tjoa@cory.Berkeley.EDU (Richard Tjoa) writes:
>>>
>>>> Ok, here's something interesting...
>>>> 9.9999...  (nine point nine repeating)  =   10  (ten)
>[...]
>
>Alright, if 9.999... = 10, then why isn't .33333... exactly 1/3?

.33333... IS exactly 1/3.

>I don't see how they are equal. I know that 9.999... with infinite 9's
>is very close to 10, but that .0000...0001 is what you need to make
>exactly 10.
>
>> 
>>     - What exactly is a non-terminating decimal?
>
>Let's keep it simple - we're talking about recurring 9's. [...]
>
>Adam Eberbach      t9114145@phillip.edu.au          fnord
>Computer Science student, Phillip Institute of Technology.

You really haven't faced up to the problem.  How many recurring 9's?
An infinite number, right?  How many zeros do you have to write in
.0000...0001 in order to make it exactly equal to the difference
between 9.9999... and 10?  An infinite number?  How do you propose to
write a 1 digit AFTER the infinite number of zeros?  There is no
"infinity-th" decimal place!

It is true that 9.9999...9 != 10 for any FINITE number of 9's after
the decimal point.  But we are talking about an INFINITE number of
9's, which is a completely different kettle of fish.  (Plenty of facts
that are true for all finite numbers are false for an infinite
number.)

The standard--and only sensible--way to interpret a non-terminating
decimal is to consider the decimal digits to be the numerators of the
terms in an infinite series, where the denominators are the successive
powers of ten.  For 9.999... we interpret the dots to mean that all
the digits are 9, in which case the series is

9.999... = 9 + 9/10 + 9/100 + 9/1000 + 9/10000 + ...
         = sum_n=0^infinity (9/10^n)

Any book on infinite series will tell you that the sum of the above
series is 10.  Not "close to 10," but EXACTLY 10.  The sum of the
infinite series is defined as the limit of the sum of the first k
terms, as k goes to infinity.

But enough pedantry.  I'm a bit surprised about your math instructor,
but maybe you didn't ask the right question.  (I could also understand
if your instructor refused to recognize 9.9999... as ANY number on the
grounds that it is improperly constructed, but that's another story.)

-- David Karr (karr@cs.cornell.edu)
From: phaedrus@cs.washington.edu (Mark Phaedrus)
Newsgroups: rec.puzzles
Subject: Re: Easy prob. question
Date: 25 Jul 92 02:25:53 GMT
Organization: University of Washington Computer Science

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.
-- 
\ Mark Phaedrus, Computer Science Major, Univ. of Washington, Seattle, WA \*\
*\ Work: phaedrus@cs.washington.edu        Play: phaedrus@u.washington.edu \*\
\*\ Cute .sig quote still under alpha-testing; planned for release in 1993. \*\
 - Hi!  I'm an anti-virus utility!  Install me in your .signature right away! -
From: imc@prg.ox.ac.uk (Ian Collier)
Newsgroups: rec.puzzles
Subject: Re: Plane Division
Date: 23 Jul 92 12:24:38 GMT
Organization: Oxford University Computing Laboratory, UK
X-Local-Date: Thursday, 23st July 1992 at 1:23pm BST

In article <54775@mentor.cc.purdue.edu>, ags@seaman.cc.purdue.edu (Dave Seaman) wrote:

>                                                        if you consider  
>what happens to the family of bisecting lines as the direction changes,  
>you will notice that you start out facing east with all the points on your  
>left, and after making a half rotation you are facing west with all the  
>points on your right. At some stage during the turning process, you must  
>have had an equal number of points on each side of the line.

As several people have now posted counterexamples to the puzzle, you may
notice that the above fails if it happens that a particular one of the
bisecting lines passes through three of the points (or two, if the rules do
not permit an answer in which the line passes through any points).

As it happens, we can easily find a sufficient condition for the answer to
be "yes", and for which the proof above holds.  Assuming that lines may pass
through points, if no three of the points are collinear then the answer is
"yes".  If the area has a centroid through which all bisecting lines pass as
long as they intersect the area (ObPuzzle: when is this true?) then
restricting the points so that no three are collinear with the centroid will
work.  If lines may not pass through points, then restricting the points so
that no two are collinear with the centroid will work.

Ian Collier
Ian.Collier@prg.ox.ac.uk | imc@ecs.ox.ac.uk
Newsgroups: rec.puzzles
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Subject: Re: 3 puzzles
Reply-To: carl@SOL1.GPS.CALTECH.EDU
Organization: HST Wide Field/Planetary Camera
Distribution: rec.puzzles
Date: Thu, 23 Jul 1992 21:25:22 GMT

In article <5071@transfer.stratus.com>, markc@eomer.hw.stratus.com (Mark Christensen) writes:
>Here are three puzzles:

Here are some spoilers:

>[1]
>	What is the first number, when spelled out (e.g. O-N-E) is
>in alphabetical order?

Ein?

>[2]
>	A man walks into a hardware store and goes to the clerk
>and says he wants an item.  "I'll take one", he says. The clerk
>says that that will be $1.  "How much for 12", the man asks?
>That will be $2 the clerk respond.  "OK, I'll take 125", the
>man says, to which the clerk responds, "That will be $3 please".
>What is the item that the man is buying.

Those little numbers you stick on the front of your house to indicate the
address.

>[3]
>  A king decided to have some fun one night and summoned three
>prisoners to be brought up from the dungeon.  They were brought
>before him and the king made an offer:
>
>"I have a box with five hats in it", he started.  "There are
>three red hats and two white hats. You will be blindfolded
>and asked to take one hat from the box.  At that time, each
>one of you will be unblindfolded and shown the hats of the other
>two prisoners (but not your own).  Then you will be asked to
>identify the color of your hat.  If you are wrong, your head
>will get chopped off.  If you don't know, back to the
>dungeons with you.  If you guess correctly, I will set you free."
>
>(Assume these prisoners are somewhat intelligent and give
>correct answers.)
>
>The first prisoner took his blindfold off, looked at the other
>two prisoner's hats and said, "I can't tell what color my hat is".
>
>The second prisoner also responded as the first prisoner.
>
>The third prisoner, who was blind, said, "I know exactly
>what color my hat is".
>
>What color is the third prisoner's hat and why can't it be the
>opposite color?

Either the second or third prisoner (or both) is wearing a red hat, since, if
they were both wearing white hats, the first prisoner would've know that he was
wearing a red hat.  Now, if the third prisoner were wearing a white hat, then
the second prisoner would've known that he had to be wearing a red hat. 
Therefore the third prisoner has to be wearing a red hat.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.
Newsgroups: rec.puzzles
From: dfs@doe.carleton.ca (David F. Skoll)
Subject: Re: Ten equals 9.9999....
Keywords: nine ten repeating digits
Organization: Dept. of Electronics, Carleton University
Date: Wed, 29 Jul 1992 13:58:59 GMT

In <1992Jul28.175614.18586@hellgate.utah.edu>
tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:

>In any event, let us suppose that for the moment, 10 does equal 9.999...
>Then by induction, 10 also equals 9.999.....(infinity)...8
>Then by induction, 10 also equals 9.999.....(infinity)...7
>Then by induction, 10 also equals 9.999.....(infinity)...6

>If we follow this through, than 10 = 9.  Then 10 = whatever we want.
>That is clearly silly, so I am inclined to agree that 10 != 9.99999....

Well, you're wrong.

9.9999... is by definition equal to 10, because a repeating decimal is
defined as the limit of the sequences of partial sums.  If you don't like
the definition, well... complain to your local mathemetician. :-)

(Anyway, your "induction" step is nonsensical, because it is impossible
to write or define 9.99999 ... 8 in any sensible way.  Perhaps you'd care to
give us a rigorous mathematical definition of just what is meant by
9.99999 ... 8?)

--
David F. SKoll
Newsgroups: rec.puzzles
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Subject: Re: carrying the torch
News-Software: VAX/VMS VNEWS 1.41    
Organization: Texas A&M University, Academic Computing Services
Distribution: rec
Date: Mon, 27 Jul 1992 04:34:00 GMT

*** SPOILER ***
















I write...
>>Set them to 'on' 'off' and 'hot'.

carl@SOL1.GPS.CALTECH.EDU writes...
>Close.
>Turn on switches 1 and 2 and turn switch 3 off.  Leave them in that state for
>several minutes.  Turn off switch 2.  Go into the other room.  The light that
>is on is controlled by switch 1.  The light that is off and warm is controlled
>by switch 2.  The light that is off and cold is controlled by switch 3.

Oh, sorry... Make that 'on' 'warm' and 'off'  ;-)

C'mon now. What did you _think_ I meant?

Dave Ring
dwr2560@zeus.tamu.edu
From: obrien@netcom.com (Robert O'Brien)
Newsgroups: rec.puzzles
Subject: Re: 3 puzzles
Date: 24 Jul 92 08:17:51 GMT
Distribution: rec.puzzles
Organization: Netcom - Online Communication Services  (408 241-9760 guest)

>>[1]
>>	What is the first number, when spelled out (e.g. O-N-E) is
>>in alphabetical order?

>forty

*I* submit that the question failed to specify 
"in ascending alphabetical order"
    ^^^^^^^^^-

Thus,  O-N-E

I thought I'd wait for the dust to settle a bit first B-)
Newsgroups: rec.puzzles
From: jimhsu@athena.mit.edu (Jimmy M.D. Hsu)
Subject: Something to look at.
Organization: Massachusetts Institute of Technology
Date: Tue, 28 Jul 1992 04:09:23 GMT

I began reading this newsgroup only today, but
I'll try my hand at spinning puzzles.  I made
this one up myself, but it's based on a game
played by the Maori that I call the "putahi" game.
Probably has a simple answer, but at least
it's unique as far as I can tell.  Here goes:

Suppose you had a pattern like this:

	O   O
     O         O
          O
     O         O
        O   O

With nine points, one in the middle.
How many different ways are there
of placing four red buttons, four
blue buttons, and one white button
on the nine spots.  If you can rotate
the pattern to get a new one, it does
not count as "different."  Is there
a generic formula for x red, y blue,
and z white with x+y+z=9?  How about
for more points in the circle?


--
James Amdee  MIT
"Drum Solo!"  -Weird Al
Newsgroups: rec.puzzles
From: jchamber@ecst.csuchico.edu (John William Chamberlain)
Subject: Re: Ten equals 9.9999....
Organization: California State University, Chico
Date: Wed, 29 Jul 1992 17:42:07 GMT

From my college calculus book:

 .3333... = 3     3      3       3        3 
            --- + ---- + ----- + ------ + ------- .......
            10    100    1000    10000    100000

      S1  = 3      ,  S2 = 3     , Sn = 3
            --             --2          --n
            10             10^          10^


 Sum from 0 to infinity of 3  is 1/3
                           --n
                           10^

From my own ideas
  
       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

Thanks for your time
  John

The book is : Calculus and Analytic Geometry
              By : Thomas and Finney 7th edition
              pp: 647-648

.
Newsgroups: rec.puzzles
From: jimhsu@athena.mit.edu (Jimmy M.D. Hsu)
Subject: Another brain teaser
Organization: Massachusetts Institute of Technology
Date: Tue, 28 Jul 1992 21:02:07 GMT

I like games with a bit of probability involved,
as you might guess.  The maori game deals with
combinations and permutations.  Here is another
game that I made up, completely original, so
don't copy unless you give me credit.  I do have
the solution to this one, if you get stuck, but
I'd rather you puzzle it out than have me post 
it since the solution is rather long.
Post comments (too simple?  too hard?)
and solutions!  But please read this carefully
and think carefully before posting solutions or
questions.


"The Killing Joke"

	Brian is playing a game with his friend,
Sam.  They use a complete poker deck with the two
Jokers.  Sam shuffles the deck thoroughly and
spreads all the cards out in one pile.  "This is
pile A," says Sam.  "We take turns putting cards
in pile B, which is the garbage pile and which is
currently empty."  Sam points to an empty spot on
the table.
	"To get cards, you can take as many as
you like from anywhere in pile A, but never from
pile B," Sam continued.  "You then discard one of
your cards into pile B and it's the next player's
turn.  The object of the game is to avoid picking
up the Jokers in pile A.  If anyone takes a Joker,
that person loses and we start a new round.
Whoever loses ten times is out of the game."
	Brian decides to think about the game to
determine the best strategy.  Since the
probability is lower in the beginning and nearly
certain towards the end when few cards remain in
pile A, Brian decides to start his rounds by
taking 27 cards, which is half of pile A.  Then,
he reasoned, he will have gotten all the low
probability cards.  He will then play his cards
and force his opponent to take the Joker in the
remaining cards.
	Sam, being a good friend, points out that
taking half the deck results in a 75% chance of
immediate loss, since there are two Jokers in the
deck.  He says his strategy is to take about six
cards at the beginning, to build up a reservoir
of cards.  He then uses all of his reservoir when
around 13 cards have been played.

	Is there a flaw in Brian's reasoning?  If
so, where?  What is the best strategy?

Variations to come...

--
James Amdee  MIT
The epitome of epitasis is when one's
epithet becomes one's epitaph.
(try saying that five times fast!)
From: ajtd@st-andrews.ac.uk (Antony J. T. Davie)
Newsgroups: rec.puzzles
Subject: number construction
Date: 29 Jul 92 16:23:49 GMT
Organization: St. Andrews University, Scotland.

Can you make the following numbers using 1,2,3,4 at most once and any
of the following operations as many times as you like:

+,-,*(times),/,^(raise to the power),!(factorial),square root.

You are also allowed to concatenate original digits to each other(2++4=24)
but not to partially formed sums so (2+4)++1 does not make 61.
You are also allowed to add decimal points and repeating points so for
                             .
example 2/.1 makes 20 and 2/.1 makes 18.

In particular you CAN'T use floor, ceiling, successor.

OK that's the rules. For easy practice here are some examples:

49 = 4!*2+1
21 = 21 (2++1)
91 = 23*4-1

Now make 89 and 137.

Tony Davie
Newsgroups: rec.puzzles
From: russotto@eng.umd.edu (Matthew T. Russotto)
Subject: Re: Ten equals 9.9999....
Date: Thu, 30 Jul 92 01:16:38 GMT
Organization: College of Engineering, University of Maryland, College Park

In article <1992Jul27.150032.12515@phillip.edu.au> t9114145@phillip.edu.au writes:

>Alright, if 9.999... = 10, then why isn't .33333... exactly 1/3?

It is.
-- 
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)