HDTenEightyP/NetNews huggingface dataset:
From: c60b-1bx@e260-3e.berkeley.edu (John S. Byun)
Newsgroups: rec.puzzles
Subject: Re: The Philosopher's Puzzle...
Date: 14 Apr 92 00:50:16 GMT
Article-I.D.: agate.sda88INN95o
Organization: University of California, Berkeley
NNTP-Posting-Host: e260-3e.berkeley.edu

In article <1992Apr9.122034@IASTATE.EDU> macster@IASTATE.EDU (Michael C Mccarty) writes:
>	Imagine you are one of three philosophers sitting under a tree in the middle of
>a nice meadow.  All of you fall asleep under the tree.  Meanwhile a group of
>birds come to perch in the tree while you are sleeping.  The birds then proceed
>to sh*t on all three of your heads.  You all wake up simultaneously and, while
>none of you can feel the sh*t on top of your head, you all burst out laughing
>because you see sh*t on the others heads.  Now assuming that none of you noticed
>who the others were looking at when they started laughing, and that none of you
>bother to feel your head (pretty big assumptions, I know, but please, bear with
>me), how can you determine that you have sh*t on your head?
>

Hmmm, this one had me thinking for a while, but I think I have the solution
worked out.  I am also going to assume that if any of the guys realized they
had sh*t on their head, they would have said so immediately, which none of
them did.

Well you know that each person is laughing because they see at least one 
other guy with sh*t on their head.  What are the two choices for yourself, 
you either have sh*t on your head or you don't.  

Let's assume that you don't have sh*t on your head.  Then take
the guy to your left for example.  What is he thinking?  He sees you who 
has no sh*t on his head and sees the other guy who has sh*t on his head.  So,
he is thinking to himself, "Hmmm, do I have sh*t on my head too?"  He 
realizes that if he didn't have any on his head, then the third guy would 
have no reason to be laughing, therefore he concludes that he does have 
sh*t on his head also.

But we know that he didn't say anything, therefore our original assumption
that you yourself has no sh*t is in error, and must conclude that you yourself
also has sh*t on your head.  He never figures out that he has sh*t on his head
because all three of you have sh*t on your heads.
Newsgroups: rec.puzzles,soc.culture.indian
From: russotto@eng.umd.edu (Matthew T. Russotto)
Subject: Re: A (Not So Obviously) simple number puzzle
Date: Wed, 08 Apr 92 17:13:58 GMT
Organization: College of Engineering, University of Maryland, College Park

In article <1992Apr8.024208.25988@CSD-NewsHost.Stanford.EDU> amorgan@Xenon.Stanford.EDU (Crunchy Frog) writes:
>
>For that matter, why go about it in such a round about fashion?  Why not
>say "Could the child be 1?  Nope, Dad would be 1 too.  Could the child
>be 2?  Unlikely, Dad would be 8.  Could the child be 3?  Sure, that would
>make Dad 3**3 or 27 and Mom would be 24."?


Arghh, I have seen some of the most atrocious math error in this
problem...

2^2 = 4, not 8.

4^4 = 256, not 64 (as another poster seemed to imply)



-- 
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)
From: bmesmer@occs.cs.oberlin.edu (Carl Muckenhoupt)
Newsgroups: rec.puzzles
Subject: Re: Where's the rest of these riddles?
Date: 9 Apr 92 19:35:00 GMT
Article-I.D.: occs.BMESMER.92Apr9143500
	<72021@netnews.upenn.edu> <1992Mar31.234116.1@amherst.edu>
Organization: Oberlin College Computer Science

In article <1992Mar31.234116.1@amherst.edu> djweisberger@amherst.edu (David Weisberger) writes:
   In article <72021@netnews.upenn.edu>, weemba@libra.wistar.upenn.edu (Matthew P Wiener) writes:
   > In article <1992Mar31.112805.7049@Miles.COM>, sjm@Miles (Scott Mark) writes:
   >>The problem you are experiencing lies in your belief that these are parts
   >>of two separate questions.
   > 
   >>The real reason that pigs fail algebra is because the buffalos live there.
   > 
   > OK, so why does a mouse when it spins?
   > -- 
   The higher, the fewer.

   So what is the difference between a duck?

One of its legs is both the same.
Now: Why does a chicken?
--

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
Newsgroups: rec.puzzles
From: tolman%asylum.utah.edu@cs.utah.edu (Kenneth Tolman)
Subject: Re: Weird puzzles just for a diversion
Date: 14 Apr 92 22:07:14 MDT
Keywords: probability,infinity
Organization: University of Utah CS Dept

In article <1992Apr14.230144.14020@EE.Stanford.EDU> quinn@EE-CF.Stanford.EDU (Christine A. Quinn) writes:
>Hi puzzlers.
>
>Here are some really weird puzzles.  

Hey EXCELLENT puzzles.  I will concentrate on 3


>Puzzle3:  Are there an infinite number of songs
>that can be written? in a paticular style (e.g. rock-n-roll)?
>of a certain length?

I was wondering, what PERCENTAGE of possible songs of each genre are already
out there?

To get a good guess of the possible songs, figure out the average length,
and then figure out how much CD memory this thing will take.  This would
give you an upper bound.  Anybody know CD memory for musical disks?

Second, we could consider a song as a set of notes played in order.  Only
the relative position will count here, as a change of key does not.  From
each note we could go to (say) 30 new notes.  Each note could be an 1/4
and combine these for wholes etc.  For a song of 50 measures, we would have
30^(16*50)  Pretty damn big.  This is a piss poor lower bound, for most of
these wont have harmony.  However, I can justify it because:

A "remake" with a new voice/instruments is basically a new song.

It would be interesting to try to perform algorithmic compression on pop
songs today, then run this filter backwards to generate music.
Newsgroups: rec.puzzles
Subject: Re: trvia - SPOILER
From: chriscasey@otago.ac.nz
Date: 15 Apr 92 13:37:04 +1300
Organization: University of Otago, Dunedin, New Zealand

In article <1992Apr12.062432.20383@rice.edu>, jwd2@elf.owlnet.rice.edu (Joseph William DeVincentis) writes:
> 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.
> 
> Well, I think he meant that there was a window in every wall.
> But, your solution can still work if you allow only one wall.
> Make your house round!
> 
Even better, build the house centred (not centered!! :-) ) on the North pole -
all directions are south!

Chris Casey
Antipodean brains are better! (at what?)
From: dant@logos.wr.tek.com (Dan Tilque)
Newsgroups: rec.puzzles
Subject: Re: word squares...
Date: 9 Apr 92 02:18:16 GMT
Organization: Upper Left Coast Liberation Army

abbey1@husc9.harvard.edu (Joshua Abbey) writes:
>
>	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...
>	The idea is that they should form actual words, across and down...

Yes but you don't limit where these "actual words" have to come from,
so don't be surprised when you get answers using words from web2 or
somewhere.

My answers follow the page feed...


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

All words in Webster's Ninth Collegiate Dictionary.

I couldn't find a solution to the other from only 9C, but I found a
solution using only one word from web2:

		Nunber 2:   T   A   T
		            E   G   0
		            E   E   L

---
Dan Tilque	--	dant@logos.WR.TEK.COM
Newsgroups: rec.puzzles
From: deghare@daisy.waterloo.edu (Dave Hare)
Subject: Re: Dying young
Organization: University of Waterloo
Date: Wed, 8 Apr 1992 00:39:47 GMT

In <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?

It seems to me that everyone is making some unwarranted assumptions here.
From the statement of the problem, there is no way to deduce what happens to
*everyone* if *someone* does push their button before 12:00.  Maybe everyone
is going to die anyway.  If this is the case, there is no gain to anyone for
anyone to push their button.

Secondly, it is not clear how reliable the statement of the situation is.
That is, do you know, of your own knowledge that there are 100 others in
the same situation as yourself, or is this just something you have been 
told by your tormentors?  The latter is, in fact, a very common strategy
used by anyone trying to manipulate someone else into doing something they
don't want to do (right across the spectrum from a child telling one parent
that the other parent said they could have icecream to prisoners of war
being told that certain behaviour will lead to certain positive consequences).

In any case, while I am the self-sacrificing type when it comes to a situation
where I am assured of the benefits to others, in the scenario given in this
problem I would not push the button because I would have no reason to accept
the word of my captors regarding what would happen to everyone else should I
do so.

Dave Hare
Newsgroups: rec.puzzles
From: yehuda@shuldig.huji.ac.il (Jonathan Berlinger)
Subject: Isaac Asimovs mysteries puzzle
Organization: Mail to News Gateway
Date: 6 Apr 92 08:04:47 GMT

Donn't know if anyone heard this one yet, so:

Why does>	Christmas = Halloween

Hint below:






















































Hint: look at the dates.

-- 

Jon Berlinger				"Shade and Sweet Water"
yehuda@cs.huji.ac.il
From: dant@logos.wr.tek.com (Dan Tilque)
Newsgroups: rec.puzzles
Subject: Re: Another Isaac Asimov puzzle -- geography
Date: 16 Apr 92 20:27:12 GMT
Organization: Upper Left Coast Liberation Army

ed@wente.llnl.gov (Ed Suranyi) writes:
>
>Interestingly, everybody who's gotten it remembers it from Asimov's essay. 

Everyone but one.  I've read maybe one or two issues of F&SF.  And I
didn't hear it from someone who had read it in Asimov's column.

In case you're interested, I didn't get it from Asimov's question, but
needed to read the modified form.  When I read the phrase "less than
50" the answer came to me.

---
Dan Tilque	--	dant@logos.WR.TEK.COM
From: ss8c+@andrew.cmu.edu (Simon Szykman)
Newsgroups: rec.puzzles
Subject: Re: trvia
Date: 11 Apr 92 23:17:56 GMT
Organization: Masters student, Mechanical Engineering, Carnegie Mellon, Pittsburgh, PA


>An architect was asked to design a square house with the window in each
>wall 
>having a southern exposure. How was this done?

Ask Santa.

Actually, you could do it elsewhere to, but then the windows would have
a southern exposure from the outside in, instead of from the inside out.

                    - Simon
From: rcbi110@muvms3.bitnet
Newsgroups: rec.puzzles
Subject: Are there any existing "treasure"-type puzzle books?
Date: 7 Apr 92 13:50:06 GMT
Organization: Marshall University


I'm not certain if this is the correct newsgroup, but it's the closest
I've found today :^)

Today's question:

How many of you remember the puzzle books "Treasure" or "Masqerade" ??
If you don't recognize the titles, they are the books that seem like ordinary
storybooks, but if you read between the lines, and study the pictures, there
are clues that lead to a real-life buried treasure....

Are there any books of this type currently out on the market, that have not
gone past their time-limit?

What are some titles of this kind of book other than the titles I mentioned?
Those are the only two I know of...

replies by either email or followups are ok.

((   ||  /\\ \\  / /\\  ((      R. Alan Monroe       ((  For best results, ((
 )) _// /--\\ \\/ /--\\  )) rcbi110@muvms3.wvnet.edu  )) squeeze sig from   ))
((              _       ((                           ((  the  bottom and   ((
 ))  ||  /\\  //  ||_/   ))    Pinkwater Rules!       )) flatten it as you  ))
((  _// /--\\ \\_ || \  ((            :^)            ((  go up.            ((

Newsgroups: rec.puzzles
From: surya@watson.ibm.com (Surya Mohanty)
Subject: Re: More Math Problems
Date: Thu, 09 Apr 1992 14:34:31 GMT
Reply-To: surya@rothko.watson.ibm.com (Surya Mohanty)
Disclaimer: This posting represents the poster's views, not necessarily those of IBM
Organization: IBM T.J. Watson Research Center

2222^5555 = (3mod7)^5555  = ( -2mod7)^1111
5555^2222 = (4mod7)^2222 =  (2mod7)^1111
hence 2222^5555 + 5555^2222 = (-2mod7)^1111 + (2mod7)^1111 = 0mod7
Newsgroups: rec.puzzles
From: trump@elaine14.Stanford.EDU (Michael Jason Lewis)
Subject: Re: The Foreign Counterfeiter
Organization: DSG, Stanford University
Date: 10 Apr 92 22:19:42 GMT

ss8c+@andrew.cmu.edu (Simon Szykman) writes:


>Fred is a counterfeiter in a foreign country.  He manages
>to steal two printing plates for local paper currency, for
>his "printing" shop.  The local currency is called bux, and
>he the plates he has are for m bux notes and for n bux notes
>(where m and n are integers).  He can therefore have a limitless
>supply of m and n bux notes.

>Fred also has a really bad guilt complex about printing up
>this funny money.  If he tries to buy something where he'd
>get change back, the idea of the person giving him real money
>back, when he's paying with fake money, causes him to confess.  

>Fred is also really cheap, so he would NEVER say anything like
>"keep the change."

>Therefore, when he buys things, he only buys things he can get
>for exact change.  The good thing is that everything in this
>country costs an integer amount of bux, including tax, so that 
>he can find things to buy for exact change often enough.

>1.    If and only if m and n have a certain property, there is
>a number X, such that Fred can buy ANYTHING that costs more than
>X bux.  What is that property? 

m and n must be relatively prime to one another (have no common factors).

>2.    What is X?

mn - 1, I think.  I didn't actually try to calculate this; it's just a 
guess, so this could easily be wrong.  But I don't think it is.

>OBFAQ-QUESTION-THAT-EVERYBODY-HATES-SEEING-BUT-SOMEONE-INEVITABLY-
>    POSTS-IT-EVERY-WEEK-ANYWAY:

>What is the next number in this series?

>1, 11, 21, 1211, 111221, ?

312211.

- Mike Lewis
trump@leland.stanford.edu
From: karlon@brchh572.bnr.ca (Karlon West)
Newsgroups: rec.puzzles
Subject: Re: An Easy Rope Problem
Date: 8 Apr 92 19:42:12 GMT
Reply-To: karlon@brchh572.bnr.ca (Karlon West)

In article <1992Mar29.041234.23132@bradley.bradley.edu>, latarra@buhub.bradley.edu (Lisa Ehren) writes:
|> bduggan@sdcc13.ucsd.edu (bri bri) writes:
|> > 	You are a thief.  A rich person has interwoven strands of
|> > gold within two ropes.  The ropes hang one foot apart from each
|> > other in a fifty foot high room.  They are attached firmly to the
|> > ceiling with a very strong glue.  The ropes are both fifty feet
|> > long.  The closest walls to the ropes are a long long way away.
|> > You have a pair of scissors.
|> > 	You want to steal as much of the two ropes as possible.
|> > However, if you drop from more than ten feet above the floor, you
|> > will break both of your legs.
|> 
|> I'm new to this group, so let me know if this won't work...
|> 
|> You didn't say that you were locked in or anything, so obviously you can
|> get out.  What about if you got a lighter and burned up the ropes thus b
|> burning off the glue and causing the gold to drop down...even if it did
|> melt, you could wait till it solidified and picked it up from the floor.
|> 
|> --
|> Fuzzies to you all.
|> 			-Latarra
|> 
|> If I am the only one who can be me, and I can only be me, if I am not me, 
|> then am I nothing and is no one me?
I didn't see any other responses to this puzzle, but assuming all you have
is a knife and you are locked in the room, is there an answer that will get
you all the rope.  I've come up with one that will get me all but about six
inches of rope.

Just curious.

karlon@bnr.ca
From: chris@peregrine.peregrine.com (Chris Cole)
Newsgroups: rec.puzzles
Subject: Re: Math (?) puzzle
Date: 16 Apr 92 06:41:05 GMT
Reply-To: chris@peregrine (Chris Cole)
Organization: Peregrine Systems, Inc.

In article <3i60iB2w164w@eastern.COM> brent-b@eastern.COM (Brent McNamee) writes:
[ audioactive series for third time in two weeks ]
[ just out of curiosity Brent, where did you hear this puzzle? ]
This puzzle is on the rec.puzzles FAQL (Frequently Asked Questions List).
For an index of the FAQL, send an email letter containing the single line:
send index
to netlib@peregrine.com

The index will be mailed via return email to the address in your
request's "From:" line.  If you are unsure of this address, and cannot
edit this line, then include in your message BEFORE the first "send" line
the line:

return_address <your_return_email_address>

From the FAQL, the answer is:
********
series.07.s
********
Each line is derived from the last by the transformation (for example)

... z z z x x y y y ... ->
... 3 z 2 x 3 y ...

"The Weird and Wonderful Chemistry of Audioactive Decay", J H Conway,
in _Open_Problems_in_Communication_and_Computation_ (Springer-Verlag, 1987), 
edited by T M Cover & B Gopinath.

Conway gives many results on the ultimate behavior of strings under
this transformation: for example, taking the sequence derived from 1
(or any other string except 2 2), the limit of the ratio of length of
the (n+1)th term to the length of the nth term as n->infinity is a 
fixed constant, namely

            1.30357726903429639125709911215255189073070250465940...

This number is from Ilan Vardi, "Computational Recreations in Mathematica",
Addison Wesley 1991, page 13.
From: BANDA@AUSVM1.VNET.IBM.COM ("Venu Prabhakar Banda")
Newsgroups: rec.puzzles
Subject: Math Problem --- Oops !!!
Date: 8 Apr 92 13:15:03 GMT
Article-I.D.: ucbvax.9204081711.AA04643

Aaaaaargh !!!!!, My first posting to rec.puzzles and I goofed....

Sorry about the problem, it is incorrect. 4444 ** 4444 is NOT divisible
by 9. In fact I realized after posting that finding (4444 ** 4444) mod 9 = 7
and the proof is quite trivial.

Heres another question......(Hopefully I wont make a mistake. This was
asked in some math olympiad)....

Problem: Consider the number 4444 ** 4444 (yes, again).

a) Let A represent the sum of the digits of above number.
b) Let B represent the sum of the digits of A.
c) Let C represent the sum of the digits of B.

Determine C ?.......

Cheers,

Venu
====
From: rgerbas@csws16.ic.sunysb.edu (Richard S Gerbasi)
Newsgroups: rec.puzzles
Subject: Security Locks
Date: 14 Apr 92 03:19:14 GMT
Organization: State University of New York at Stony Brook
Originator: rgerbas@csws16.ic.sunysb.edu

OK everyone, I hope this one isn't two much for this newsgroup.  
In one of Applied Math classes, we were assigned this problem as an
extra credit problem.  Due to a growing number of tests I have to take,
I figured I would post it here for some advanced insight on the problem
and perhaps a solution. By solution I mean fairly detailed so I can 
follow it.  Oh, and by the way if my Professor is reading, I promise not
to use any form of solution given to me from this Newsgroup.  (Well
maybe not promise, but I will try!!!)  Well here it is!

Many of you have used secutiy doors that have a numeric keypad on the
door to serve as a lock.  A typical such lock has 5 buttons (numbered 1
through 5) and can be assigned any "code" that obeys the following rules:

	The code consists of between 1 and five "pushes", each puch of
one or more buttons.  The buttons that are depressed during one push 
are all depressed at once; e.g., you might depress buttons "1", "3", and
"5" all at once.
	No button is used more than once in the entire code; once a button
has been depressed during some push, it will not be allowed to be used in
any future push.
	Not all buttons need to be used.

Example codes:
1,3,5
3,5,1
35,1
12345(all at once)
12,34 (which is same as 21,43)
3
1,2345

Question: How many different codes can such a lock have?  (Careful
when counting not to count an incident twice) If you could try one
code every 10 seconds how long would it take you to break in to the room?

This lovely question was brought to you by Professor Mitchell.

Good luck.

Rick   rgerbas@libserv1.ic.sunysb.edu

I do realize it would be a whole lot easier to break the door down.
-- 
Rick The Owner of "   "
Newsgroups: rec.puzzles
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Subject: Re: Math Problem
Reply-To: carl@SOL1.GPS.CALTECH.EDU
Organization: HST Wide Field/Planetary Camera
Date: Sat, 11 Apr 1992 04:28:41 GMT

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

It's not.  3 is prime and not a factor of 4444.  Therefore it cannot be a
factor of any integral power of 4444.  Therefore it is not a factor of 4444 **
4444.  Any number divisible by 9 must be divisible by 3.  Therefore 4444 **
4444 is not divisible by 9.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXes 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.
From: hxie@grad1.cis.upenn.edu (Hong Liang Xie)
Newsgroups: rec.puzzles
Subject: Re: More Math Problems
Date: 10 Apr 92 21:29:17 GMT
Organization: University of Pennsylvania

In article <1992Apr10.183159.11305@ntg.com> slandrum@ntg.com (Stephen H. Landrum) writes:
>
>Nice try, but your logic has a gap.  By your reasoning, 233 is a multiple of
>7:
>        (8^2 + 13^2) mod 7
>        = ((7*1+1)^2 + (7*1+6)^2) mod 7
>        = (1^2       +       6^2) mod 7
>        = (1+6)(...) mod 7
>        = 0
>
>Clearly this is false since 233 mod 7 = 2.

You missed the point that (x^n + y^n) = (x+y)(...) only when
n is an odd number.   

-Hong, CIS Dept, U of Penn
From: annan@vax.oxford.ac.uk
Newsgroups: rec.puzzles
Subject: Re: The Philosopher's Puzzle...
Date: 11 Apr 92 16:34:55 GMT
Organization: Oxford University VAXcluster

In article <1992Apr9.122034@IASTATE.EDU>, macster@IASTATE.EDU (Michael C Mccarty) writes:
> how can you determine that you have sh*t on your head?

1. How about asking one of the others "do I have shit on my head?".
2. Look in a mirror.
-- 
James Annan.
                [--------------------------------------]
                [  "The only good Tory is a lavatory"  ]
                [--------------------------------------]