HDTenEightyP/NetNews huggingface dataset:
From: hmasum@alfred.carleton.ca (Hassan Masum)
Subject: 'easy' math puzzles
Summary: 'easy' math puzzles
Organization: Carleton University
Date: Tue, 14 Apr 1992 05:34:55 GMT

          eAsY if you know how!            \___/



1.  A Toronto lawyer normally gets home by taking the Bloor subway
line to a station close to her house (Warden station), and then
getting picked up by her husband in their 1974 Ford station wagon
which has little bits of rust falling off the back.  This husband
manages the extraordinary feat of driving at exactly the same speed
from his house to the station and back, turning around instantaneously
and picking up his wife with a megateslamagnet (she has iron threads
woven into her dress).
          So anyway, normally the lawyer gets picked up at 17:00.  One
day, a fine sunny Tuesday except for little patches of smog caused by
large quantities of industrial pollutants being pumped into the
atmosphere (but that's another story), she gets to the station at
16:00 because she's been temporarily laid off due to a large economic
depression.  In somewhat of a depression herself, she seeks to cheer
herself up by walking back to her suburban bungalow.  On the way to
this house, walking north on the east side of Warden street, she
encounters her husband driving south to the Warden Avenue subway
station; being a keen-eyed and alert driver (as indeed one must
obviously be to travel at a constant speed (see paragraph 1, above)
through rush-hour traffic in Toronto) he notices his beloved wife,
does one of his fancy electromagnetic semidonut tricks, and reaches
home ten minutes earlier than normal.
          How long was the lawyer walking for before she was picked
up?


2.  Two infinitely long identical cylinders with finite radius
intersect at right angles, (obviously) passing completely through each
other.  What is the maximal volume of intersection in terms of the
radius of the cylinder(s)?


3.  Call a 'polycube' a union of identical cubes, joined so that
connecting faces are completely covered.  Find a polycube such that,
given an infinite number of congruent copies of the polycube, these
copies may be put together, with no interior hole, so that each pair
shares some two-dimensional boundary.
          (If you find the polycube, you will have shown that, while 4
colors suffice to make a 2D map, a 3D 'map' requires an infinite
number.) 


4.  Show that, for every prime p > 5, there is some number b
consisting of a string of n ones (called an n-repunit) such that p
divides b.  Give an explicit formula for n in terms of p (for one
b--there may of course be (many) more than one for any p).


5.  Is it possible to regularly tile a plane with a pattern having
pentagonal symmetry?
Newsgroups: rec.puzzles
From: wyg9633@uxa.cso.uiuc.edu (Wooed Sun)
Subject: Re: Ladder puzzle
Organization: University of Illinois at Urbana
Date: Fri, 17 Apr 1992 18:58:14 GMT

holt@lambda.msfc.nasa.gov (Tom Holt) writes:


> I have an interesting little puzzle.

>     -   |       |
>     |   |\     /|  -
>     |   | \   / |  |
>         |  \ /  |
>    15   |   x   |  10
>         |  / \  |
>     |   | / 6 \ |  |
>     |   |/  |  \|  |
>     -   ---------  -
            3 : 2

>         |-  ?  -|


> Two ladders cross one another in an alley. Their bottom edge is 
> butted into the corner. The first one hits the far wall 15 ft. up.
> The second one hits the other wall 10 ft. up. The point at which 
> they intersect is 6 ft. high. What is the width of the alley?

Any value larger than 0.

> I'd like to see how a solution is reached, also.


-- 
email to : w-yang@uiuc.edu (wyg9633@uxa.cso.uiuc.edu)
Newsgroups: rec.puzzles
From: walsh@spider.wri.com (Jerry Walsh)
Subject: Re: Math Problem (Solution?)
Organization: Wolfram Research, Inc.
Date: Thu, 16 Apr 1992 00:15:08 GMT

In article <1992Apr8.042357@IASTATE.EDU> macster@IASTATE.EDU (Michael C Mccarty) writes:
>	An interestion point was just brought up... Does the number have to be in base
>10?  If base 8 is used, the number is then divisible by 9:
>
>4444^4444 (base 8) = 2340^2340 (base 10)=
>((9*260)^260)^9=
>(9^260)^9*(260^260)^9
>

Since you recognized that 9 base 10 is 11 base 8, you could have spared
the calculator work and just noted that 4444 base 8 is divisible by 11
base 8 by inspection; that is, the "long" division is very quick and
easy.

4444 / 11 is 404 in any base greater than 4.

Jerome Walsh
walsh@wri.com
From: morris@anasaz (Bob Morris)
Newsgroups: rec.puzzles
Subject: Original puzzle inspired by...
Date: 16 Apr 92 16:23:17 GMT
Organization: Anasazi, Inc.  Phoenix, Az

Kenneth Tolman writes:

>Now, what I would think would be a REALLY cool puzzle is to pick five
>cities that form an even pentagram on the surface of the earth, and
>ask about these.

OK, you have inspired me.  If you covered the Earth with 12 pentagrams
(aka pentagons) forming something like a dodecahedron, how long would
each "edge" be?  (this is a little different than a real dodecahedron,
as its edges are arcs, not lines).

Similarly, if you drew one in the skies, what is the angle formed
between two adjacent corners with you at the vertex?

For extra credit, show all work and generalize to any of the various
geometric solids (tetrahedron, cube, octahedron, icosahedron)

Bob Morris
aka morris@anasazi.com  (new address)
anasaz!morris@asuvax.eas.asu.edu (old address still works)
My opinions only, of course...
Newsgroups: rec.puzzles
From: slandrum@ntg.com (Stephen H. Landrum)
Subject: Re: More Math Problems
Organization: New Technologies Group, Inc.  Palo Alto CA
Date: Fri, 10 Apr 1992 18:31:59 GMT

In article <73348@netnews.upenn.edu> hxie@grad1.cis.upenn.edu (Hong Liang Xie) writes:
>In article <9204081953.AA09310@ucbvax.Berkeley.EDU> BANDA@AUSVM1.VNET.IBM.COM ("Venu Prabhakar Banda") writes:
>>1. Show that (2222^5555 + 5555^2222) is divisible by 7?

>A series of calculations can solve this problem easily.

>(2222^5555 + 5555^2222) mod 7
>	= ((317x7+3)^5555 + (793x7+4)^2222) mod 7
>	= (       3 ^5555 +        4 ^2222) mod 7
>	= (     243 ^1111 +       16 ^1111) mod 7
>	= (( 34x7+5)^1111 + (  2x7+2)^1111) mod 7
>	= (       5 ^1111 +        2 ^1111) mod 7
>	= (5+2)(...) mod 7
>	= 0  



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.

-- 
Stephen H. Landrum                                        VOICE: (415) 813-8909
        Domain: slandrum@ntg.com           UUCP: ...netcomsv!ntg!slandrum
 USNAIL: New Technologies Group Inc. 2468 Embarcardero Way, Palo Alto CA 94303
From: koerse@fwi.uva.nl (Niels Koerse (I88))
Newsgroups: rec.puzzles
Subject: Re: Geographical puzzle.
Date: 15 Apr 92 10:27:12 GMT
Organization: FWI, University of Amsterdam

wft@math.canterbury.ac.nz (Bill Taylor) writes:

>What property do the following southern-hemisphere cities and towns have 
>in common? As far as I can check, it is an exhaustive list (for this property).

>Buenos Aires
>Melbourne
>Mar del Plata      (Argentina)
>Bahia Blanca       (Argentina)
>Christchurch       (New Zealand)   ........[ yay ! home team !]
>Comodoro Rividavia (Argentina)
>Punta Arenas       (Chile)
>Ushuaia            (Argentina)

I think something with the meaning of the names.
>Buenos Aires
 good   air  	(problem here)
>Melbourne
 		never understood the australians :)
>Mar del Plata      (Argentina)
 sea ??  beach
>Bahia Blanca       (Argentina)
 white beach
>Christchurch       (New Zealand)   ........[ yay ! home team !]
 christian church
>Comodoro Rividavia (Argentina)
 yes i should train my spanish or is it portoguese

So i think they all say things in a that are/were obvious.

>There is a similar list for the Northern hemisphere. I'll leave it to
>some native to work that one out.
naah

>---------------------------------------------------------------------
>  Bill Taylor                             wft@math.canterbury.ac.nz 
>---------------------------------------------------------------------
>               " Pretentious ?  MOI !? "
>---------------------------------------------------------------------


****************************** koerse@fwi.uva.nl ******************************
******************************** .sig ** naaah ********************************
-- 
****************************** koerse@fwi.uva.nl ******************************
******************************** .sig ** naaah ********************************
From: morris@anasaz (Bob Morris)
Newsgroups: rec.puzzles
Subject: Re: Weird puzzles just for a diversion
Keywords: probability,infinity
Date: 16 Apr 92 16:48:53 GMT
Organization: Anasazi, Inc.  Phoenix, Az

In article <1992Apr14.230144.14020@EE.Stanford.EDU> quinn@EE-CF.Stanford.EDU (Christine A. Quinn) writes:
>
>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?
>
SPOILERS ALWAYS SEEM TO FOLLOW ^L, DON'T THEY?

I'd say no.  If the notes are of finite duration and fit into the
standard notation (half, quarter, etc.), the number of songs of
of length N seconds using M different tones with the shortest note
being duration D seconds is given by:

	M ^ (N/D)

Just multiply by the number of voices (aka instruments) for a band,
choir, or orchestra and it's still finite.  Since the number of songs
of a particular style is a subset of the total, it is finite also.

This can be a pretty large number.  For example, for quarter notes of
1/4 second using only 10 different notes, there are 10^480 different
2 minute songs.  (This exceeds the highest estimates of the number of
subatomic particles in the universe, but it's still finite.)

If, on the other hand, you allow a continuous scale of tones, M is
infinite and the answer is yes.  Also, if you allow a continuous variation
in the duration of each note, the number of songs is infinite.

If, on the third hand, time and space are quantized (we're beyond the
farthest reaches of physics now), the number of *everything* is finite.

Bob Morris
aka morris@anasazi.com  (new address)
anasaz!morris@asuvax.eas.asu.edu (old address still works)
My opinions only, of course...
Newsgroups: rec.puzzles
From: 082589@hyperion.lanl.gov (Jerry Halladay P350/7-5123 P350)
Subject: Re: Where's the rest of these riddles?
Reply-To: 082589@hyperion.lanl.gov
Organization: Los Alamos National Laboratory
Date: Fri, 10 Apr 1992 17:23:09 GMT

Whats the difference between a duck? ... one leg is both the same!
From: kontos@clipper.ingr.com (Thorne Kontos)
Newsgroups: rec.puzzles
Subject: A question of addresses
Keywords: A simple puzzle for Wednesday
Date: 8 Apr 92 20:25:56 GMT
Distribution: na
Organization: Intergraph Advanced Processor Division - Palo Alto, CA

Sally and Sue have a strong desire to date Sam. They all live on the
same street yet neither Sally or Sue know where Sam lives. The houses
on this street are numbered 1 to 99.

Sally asks Sam "Is your house number a perfect square?". He answers.
Then Sally asks "Is is greater than 50?". He answers again.

Sally thinks she now knows the address of Sam's house and decides to
visit. When she gets there, she finds out she is wrong. This is not
surprising, considering Sam answered only the second question
truthfully.

Sue, unaware of Sally's conversation, asks Sam two questions.
Sue asks "Is your house number a perfect cube?". He answers.
She then asks "Is it greater than 25?". He answers again.

Sue thinks she knows where Sam lives and decides to pay him a visit.
She too is mistaken as Sam once again answered only the second 
question truthfully.

If I tell you that Sam's number is less than Sue's or Sally's,
and that the sum of their numbersis a perfect square multiplied
by two, you should be able to figure out where all three of them
live.
From: bas@phys.uva.nl (Bas de Bakker)
Newsgroups: rec.puzzles
Subject: Re: A Modest Proposal
Date: 13 Apr 92 15:49:46 GMT
Organization: Institute for Theoretical Physics, Amsterdam, the Netherlands
In-reply-to: tvaughan@central.rutgers.edu's message of 14 Apr 92 13:43:08 GMTLines: 19

>>>>> On 14 Apr 92 13:43:08 GMT, tvaughan@central.rutgers.edu (Timothy E. Vaughan) said:

tev>Despite the Subject of this posting, it is not intended to be
tev>sarcastic.

Neither is this one.

tev>I propose that people who know the answer to a posted puzzle simply
tev>because they heard it before (frequently LONG before) refrain from
tev>posting an answer.  People who genuinely FIGURED OUT a puzzle that was
tev>NEW TO THEM should happily post.  I believe the savings in bandwidth
tev>would be astounding.

It would save even more bandwidth and annoyance if people wouldn't
post puzzles that so many other people have heard LONG before.

tev>Tim

Bas.
Newsgroups: rec.puzzles
From: guckes@math.fu-berlin.de (Sven Guckes)
Subject: Re: Math Problem
Originator: guckes@medusa
Organization: Free University of Berlin, Germany
Date: Tue, 7 Apr 1992 19:59:12 GMT

BANDA@AUSVM1.VNET.IBM.COM ("Venu Prabhakar Banda") writes:

>Problem: Show that (4444 ** 4444) is divisible by 9
>
>Cheers,
>
>Venu
>====

	4444 * 4444 = 6583045 * 3 + 1

	Boring puzzle. ;)
From: ecarsten@next1.cs.umr.edu (Edward Carstens)
Newsgroups: rec.puzzles
Subject: square 1
Date: 7 Apr 92 19:54:04 GMT


Thanks to whomever it was that responded.  I haven't had time to play with it  
lately but I thank you for your suggestion and I'll try it next time I play  
with square 1.

-Ed Carstens
Creator of JugglePro
From: ss8c+@andrew.cmu.edu (Simon Szykman)
Newsgroups: rec.puzzles
Subject: Re: Math (?) puzzle (suggestion)
Date: 16 Apr 92 03:07:43 GMT
Organization: Masters student, Mechanical Engineering, Carnegie Mellon, Pittsburgh, PA


  A    RRRR   RRRR   RRRR    GGGG    GGGG   H   H  !!  !!  !!
 A A   R   R  R   R  R   R  G       G       H   H  !!  !!  !!
A   A  RRRR   RRRR   RRRR   G  GGG  G  GGG  HHHHH  !!  !!  !!
AAAAA  R  R   R  R   R  R   G   G   G   G   H   H  
A   A  R   R  R   R  R   R   GGG     GGG    H   H  !!  !!  !!


THE PUZZLE THAT WOULDN'T DIE!

Is there anyone out there who would volunteer to post this puzzle
and the spoiler once a week with a subject header something on the
order of "READ THIS IF YOU ARE NEW"?

That way, the people who have seen it before recognize it and skip
it, the people who haven't seen it before see it on rec.puzzles, 
and therefore don't post it to rec.puzzles, and best of all, we
save bandwidth since we have only one spoiler posted instead of
the usual dozen.

                                           - Simon
From: meng@idt.unit.no (Tore Mengshoel)
Newsgroups: rec.puzzles
Subject: Re: order of numbers?
Date: 9 Apr 92 15:23:50 GMT
Reply-To: meng@idt.unit.no (Tore Mengshoel)
Distribution: rec
Organization: Div. of CS & Telematics, Norwegian Institute of Technology

In article <9385@wrgate.wr.tek.com>, dant@logos.wr.tek.com (Dan Tilque) writes:
|> pfalstad@phoenix.Princeton.EDU (Paul Falstad) writes:
|> >dant@logos.wr.tek.com (Dan Tilque) wrote on 6 Apr 92 17:01:18 GMT:
|> >>> > 	2, 9, 8, 4, 5, 7, 6, 3, 1
|> >
|> >Spoilers...
|> 
|> [ stuff deleted ]
|> >>	8, 3, 1, 5, 9, 7, 6, 4, 2
|> >I'll guess German, based on where the "2" is.
|> 
|> Right again.
|> 
If this shall be german, the seven (sieben) should be after the six
(sechs), if I am not mistaken (my german is a little rusty..)

|> >>All four of these lists, 7 comes in the sixth position.  I wonder if this
|> >>is only a coincidence.
|> >
The correction destroys this regularity


|> [more deletion]
|> ---
|> Dan Tilque	--	dant@logos.WR.TEK.COM


Tore
Newsgroups: rec.puzzles
From: tolman%asylum.utah.edu@cs.utah.edu (Kenneth Tolman)
Subject: Re: Another Isaac Asimov puzzle -- geography
Date: 14 Apr 92 11:42:11 MDT
Organization: University of Utah CS Dept

In article <122473@lll-winken.LLNL.GOV> ed@wente.llnl.gov (Ed Suranyi) writes:
>Here's a geography puzzle that appeared in one of Asimov's essays in
>_The Magazine of Fantasy and Science Fiction._  I have modified it
>slightly to make it slightly easier, because as he stated it the puzzle is
>virtually impossible (IMHO).  But I'll give Asimov's version first,
>then mine, so people can start with his alone.  Even my version is
>difficult enough so that I'd be surprised if anyone who hasn't seen
>or heard of Asimov's solution could get it.  In fact, I think it's 
>the most difficult geography trivia question I've ever seen!
>
>Here's Asimov's version:  What do New York City, and Xenia, Ohio have
>in common?
>
>My version is below, for those who don't want to see it:
>


[spoiler?] NOT



































These cities have some wierd latitude or longitude.  Some possible
permutations on this are:

This city has the same approximate lat as long.
All these cities lie along the same latitude.
These cities have an official population which is equivalent to a constant
times their latitude.
Take the letters of their names, add them up, and find the latitude.

Now, what I would think would be a REALLY cool puzzle is to pick five
cities that form an even pentagram on the surface of the earth, and
ask about these.

Newsgroups: rec.puzzles
From: jmarlan@titan.nmt.edu (Jon Marlan)
Subject: Re: farms
Organization: New Mexico Institute of Mining and Technology
Date: Wed, 8 Apr 92 19:07:37 GMT


In article <emulcahy.702742368@unix2.tcd.ie> emulcahy@unix2.tcd.ie (QUETZALCOATL              MONTEZUMA OAXACA TENOCHTITLAN INTI RAYNI RA CHICHEN ITZA TLAXCALA MACCHU PICCHU CUZCO AKRANES REKJAVIK AKUYEYRI KEFLAVIK HAFNAFJIORDHUR GODTHAAB THULE MINNESOTA NORTHSTARS SEATTLE SEAHAWKS KON-TIKI) writes:
>
>     
>
>
>    If a Horse costs #10,
>
>    a goat costs &1,
>
>    & 8 chickens cost #1,
>
>    
>    how many of each animal must you buy to spend exactly #100,
>
>    & have 100 animals ,with at least one of each kind?

The trick to this one is realizing that the total number of 
chickens and horses must be the same as their combined costs.

7 horses at #70, 21 goats at #21, and 72 chickens at #9

Total number of animals = 7 + 21 + 72 = 100
Total cost of animals = #70 + #21 + #9 = #100


Jon

---------------------------------------------------
"You are what goes into your mind" -- Zig Ziglar
---------------------------------------------------







Newsgroups: rec.puzzles
From: amorgan@Xenon.Stanford.EDU (Crunchy Frog)
Subject: Re: Brick Problem (crude ASCII)
Keywords: brick, graphical, puzzle
Organization: Computer Science Department, Stanford University.
Date: Mon, 6 Apr 1992 17:53:19 GMT

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

>Here's one I've been toying around with while bored in class for
>awhile, but I've ony managed to solve one of them.  An uncle described
>these to me as being solvable, so don't give up... (that is, if you
>believe my uncle. :)
>I myself have only been able to solve one of them.  I won't tell you which.

My psychic powers tell me it is the first one.  Eatcher heart out James
Randi.

>Ok, here, in crude ascii, is a picture of a brick wall, with three
>bricks on top, two on the middle row, and three again on the bottom
>row.
>The object is to draw a curvilinear line through every wall of every
>brick, without going through any given wall more than *once*.
>Note:  Bricks in the middle row have *six* walls, ok?  I can't think
>of a better way to describe this.  Center bricks on the top and bottom
>rows have *five* walls.
>
>  |----|-----|----|
>  |    |     |    |
>  |----|--|--|----|
>  |       |       |
>  |----|--|--|----|
>  |    |     |    |
>  |----|-----|----|

.. deleted ..

>Same problem, different brick layout...
>
>
>  |-------|-------|
>  |       |       |
>  |----|--|--|----|
>  |    |     |    |
>  |----|-----|----|
>  |       |       |
>  |-------|-------|

Bad news Paulster - your Uncle is a lying sack of mucus.  This last
one can't be solved on a planer surface (it can be solved on a genus
2 surface if the holes are in the right places).

Short proof:  If the line doesn't end in a particular room it must
exit as often as it enters - thus any room in which the line doesn't
end must have an even number of walls.  The rooms which have the endpoints
of the lines must have an odd number of walls for the same reason.  In
the previous diagram there are 4 rooms with an odd number (5) walls -
it can't be done.

>orange@ccs.carleton.ca ------- Paul Vallee ---- 542429@acadvm1.uottawa.ca
>------------Home: (613)569-1068 --------- Work: (613)995-6921------------

C "There is no Santa Claus either" Frog
Newsgroups: rec.puzzles
Subject: Re: Math Problem
From: tatsuya@hamblin.math.byu.edu (& Kawasaki)
Date: 11 Apr 92 11:56:12
Organization: Brigham Young University

In article <9204071337.AA25632@ucbvax.Berkeley.EDU> BANDA@AUSVM1.VNET.IBM.COM ("Venu Prabhakar Banda") writes:
Path: hamblin.math.byu.edu!yvax.byu.edu!byuvm!cunyvm!ukma!asuvax!cs.utexas.edu!qt.cs.utexas.edu!
    zaphod.mps.ohio-state.edu!cis.ohio-state.edu!ucbvax!AUSVM1.VNET.IBM.COM!BANDA
   Newsgroups: rec.puzzles
   From: BANDA@AUSVM1.VNET.IBM.COM ("Venu Prabhakar Banda")
   Date: 7 Apr 92 13:37:11 GMT
   Sender: daemon@ucbvax.BERKELEY.EDU
   Lines: 6

   Problem: Show that (4444 ** 4444) is divisible by 9

   Cheers,

   Venu
   ====


correction:
4444^4444 is not divisible by 9.  it will have 7 as a remeinder..
Did I miss something????
--
                                         _   /|
Tatsuya                                  \'o.O'
tatsuya@hamblin.math.byu.edu             =(___)=
EMT:901006  Ham: N7UQJ                      U
From: dant@logos.wr.tek.com (Dan Tilque)
Newsgroups: rec.puzzles
Subject: Re: order of numbers?
Date: 8 Apr 92 20:50:53 GMT
Distribution: rec
Organization: Upper Left Coast Liberation Army

pfalstad@phoenix.Princeton.EDU (Paul Falstad) writes:
>dant@logos.wr.tek.com (Dan Tilque) wrote on 7 Apr 92 02:44:39 GMT:
>>pfalstad@phoenix.Princeton.EDU (Paul Falstad) writes:
>>>The Romance languages are similar:
>>>5, 2, 8, 9, 4, 7, 6, 3, 1
>>>and
>>>5, 2, 9, 8, 4, 7, 6, 3, 1
>
>>I'm sure that neither sequence is French and
>>probably not Rumanian, but any of Italian, Spanish or Portuguese.
>
>The first one was supposed to be French, but my memory is probably wrong.
>I can't check, since I don't have a French dictionary around.

Oops.  Hold on a minute while I wipe the egg from my face.  I'd written
down the French list and alphabetized the 6 and 7 wrong.  You did give the
proper French order.

And even more embryonic avians in proximity to my anterior orifice, I
also screwed up the German list I posted (also getting the 6 and 7
switched).  Others have pointed this out.

I think I'm going to shut up on this subject while I'm behind (my egg
wiping towel is getting saturated).

---
Dan Tilque	--	dant@logos.WR.TEK.COM
From: djweisberger@amherst.edu (David Weisberger)
Newsgroups: rec.puzzles
Subject: Re: Where's the rest of these riddles?
Date: 1 Apr 92 03:41:16 GMT
Article-I.D.: amherst.1992Mar31.234116.1
Organization: Amherst College, Amherst, MA.

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?

> -Matthew P Wiener (weemba@libra.wistar.upenn.edu)


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 David Weisberger                  djweisberger@amh.amherst.edu

         "This is not a quote.  No-one ever really said this."
	      (Although Chris Someone did quote it once.)