HDTenEightyP/NetNews huggingface dataset:
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Subject: Why there are no balls left in the urn
Keywords: balls, urn, infinite
Organization: Purdue University
Date: Fri, 6 Nov 1992 18:39:16 GMT

Analysis of the Balls-in-the-Urn Problem
----------------------------------------

To make things definite, let's assume that balls 1-10 are placed
in the urn at time t = -1 (1 minute to the hour), and ball #1 is
removed. Balls 11-20 are placed in the urn at t = -1/2, and ball
#2 is removed. In general, balls #10*n-9 through 10*n are placed
in the urn at time t = -1/n, and ball #n is removed.

For each n>0, ball #n enters the urn at time t = -1/ceiling(n/10)
and leaves again at time t = -1/n, never to return.  We can describe
the behavior of ball #n for any time t by the function

		/	1, if -1/ceiling(n/10) <= t < -1/n,
	f (t) = |
	 n	\       0, otherwise.

Therefore, f (t) = 1 if ball n is in the urn at time t, and 0 otherwise.
            n

The total number of balls in the urn at time t is just

	        oo
	       ___
               \
	f(t) =  >   f (t)
	       /     n
	       ___
	       n=1

and this formula holds for all t.  In other words, we merely count
up the balls that are in the urn at time t, and let the total be
f(t). What could be simpler?

The series is absolutely convergent, because for each time t, there
are only finitely many balls in the urn. In fact, if -1/k <= t <
-1/(k+1) for some integer k > 0, then balls k+1, k+2, ..., 10*k-1,
10*k are in the urn at time t, a total of 9*k balls altogether.
For t < -1 and t >= 0 the urn is empty.  It follows that the exact
number of balls in the urn for any time t is given by

	       /       9*k, if -1/k <= t < -1/(k+1),
	f(t) = |
	       \       0,   if t < -1 or t >= 0.

Clearly, f is discontinuous, since f(0) = 0 but the left-handed
limit is +infinity.

--
Dave Seaman
ags@seaman.cc.purdue.edu
Xref: sparky rec.puzzles:7151 sci.math:14701
From: sc5x+@andrew.cmu.edu (Steven Pei-Han Chai)
Newsgroups: rec.puzzles,sci.math
Subject: Re: gays and condoms (PARTIAL SPOILER)
Date: 9 Nov 92 20:17:07 GMT
Article-I.D.: andrew.QezpDXO00aw6F_uGEn
	<Nov.7.13.49.53.1992.7608@remus.rutgers.edu>
Organization: Freshman, MCS general, Carnegie Mellon, Pittsburgh, PA

Hi!

I remember seeing a post here where there were about hundred situation
puzzles listed with the answers.  Could that person repost them for me? 
I'd like to print them out.

thanks.

Happy  (happy+@cmu.edu)
From: skataba@pbhyf.PacBell.COM (Sam Atabaki)
Newsgroups: rec.puzzles
Subject: Re: Break Time Head Work
Date: 5 Nov 92 19:59:33 GMT
Reply-To: skataba@PacBell.COM (Sam Atabaki)
Organization: Pacific * Bell, San Ramon, CA

In article <1992Nov4.133223.16332@aio.jsc.nasa.gov> nas_ps@jsc.nasa.gov writes:
>The following word problems appeared recently in the Design News:
>
>1.	A hiker can average two miles per hour uphill and six miles per hour
>	downhill.  Going uphill and down, and if s/he spends no time at the
>	summit, what will be his/her average speed for an entire trip ?
>
Let m be the distance (one-way). Then,
   it takes him m/2 hours to go up the hill
   it takes him m/12 hours to go back down the hill
So, total time spent to travel 2m miles is (m/2 + m/12)= 7m/12 hours,
thus, the average speed is:  24/7 miles per hour.

Sam Atabaki
From: dik@cwi.nl (Dik T. Winter)
Newsgroups: rec.puzzles
Subject: Re: Formulae for pi
Date: 8 Nov 92 01:35:02 GMT
Organization: CWI, Amsterdam

If you check sci.math about one or two weeks ago there was an article
that gave a (short) formula wich was a correct approximation of pi in
42 billion digits.  You could try that one.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl
Newsgroups: rec.puzzles
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Subject: Re: Formula for Pi
Organization: Purdue University
Date: Tue, 10 Nov 1992 14:41:54 GMT

In article <1992Nov9.162357.10018@vax.oxford.ac.uk> peake@vax.oxford.ac.uk  
writes:
> One formula which you might use, after complicated programming, is
> 
> Pi = 16*arctan(5)-4*arctan(1/239)
>                      
> and then use arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + ...


That should be

  Pi = 16*arctan(1/5)-4*arctan(1/239)

--
Dave Seaman
ags@seaman.cc.purdue.edu
From: radcliff@csd4.csd.uwm.edu (David G Radcliffe)
Newsgroups: rec.puzzles
Subject: Re: Cliff Puzzle 18: Spoiler
Date: 12 Nov 1992 17:46:38 GMT
Organization: UW-Milwaukee Math Department
NNTP-Posting-Host: 129.89.7.4

The problem is to simplify 
  sqrt( 1 + x sqrt( 1 + (x+1) sqrt ( 1 + (x+2) sqrt ( (x+3) sqrt ( 1 + . . .
when x > 0.

The answer is x + 1.  Consider the following identities:

  x + 1 = sqrt( 1 + x (x + 2) )
  x + 2 = sqrt( 1 + (x + 1) (x + 3) )
  x + 3 = sqrt( 1 + (x + 2) (x + 4) )
  x + 4 = sqrt( 1 + (x + 3) (x + 5) )
        .
	.
	.

Then perform an "infinite substitution" to get the required result.
(Of course, convergence should be proved as well.)

Can anybody simplify the following nested radical?

  sqrt(1 + sqrt(2 + sqrt(3 + sqrt(4 + . . .

--
David Radcliffe                                
radcliff@csd4.csd.uwm.edu
From: dcarroll@oucsace.cs.ohiou.edu (Dana Carroll)
Newsgroups: rec.puzzles
Subject: Re: Logic Puzzle
Date: 6 Nov 92 16:04:49 GMT
Organization: Ohio University CS Dept,. Athens

In article <721041158.AA07433@csource.oz.au> Tim.Epstein@f526.n635.z3.fidonet.org (Tim Epstein) writes:
> 
>Which of the following 5 sentences are true?
> 
>a) It is not the case that 2 consecutive sentences are both false. 
>b) There are fewer false than true sentences. 
>c) It is not the case that 3 consecutive sentences are all false.
>d) It is not the case that 2 consecutive sentences are both true. 
>e) There are exactly 3 false sentences. 
> 

This puzzle is not solvable!  My reasoning is as follows:

1) Assume (a) is false.  If (b) is true, then either 
   1.1 (c) and (d) are false, which contradicts (b), or
   1.2 (d) and (e) are false, which also contradicts (b).

2) If (a) is false, then (b) must also be false.  But then
   2.1 one of (c), (d), (e) must be false.
       2.11 assume (e) false.  Then one of (c), (d) also false.  But
            2.111 (c) false implies (d) true, but then (c) is true.
            2.112 (d) false is an immediate contradiction.
       2.12 then (e) must be true.
            2.121 (c) false implies (d) true.  (e) also true contradicts (d)
            2.122 (d) false is an immediate contradiction.

3) So, (a) must be true.  Then either (b) is True of false.
   3.1 (b) true.  Then two of (c), (d), (e) alse true.
       3.11 (c), (d) true contradicts (d)
       3.12 (c), (e) true contradicts (e)
       3.13 (d), (e) true also contradicts (e)
   3.2 So (b) must be false.  Then at least two of (c), (d), (e) also false.
       3.21 (e) false implies (c), (d) false contradicting (c)
       3.22 So (e) true and thus (c), (d) false, contradicting (e)!


So (a) can be neither true nor false!

Did I do something wrong?
--Dana
From: elkies@ramanujan.harvard.edu (Noam Elkies)
Newsgroups: rec.puzzles
Subject: Re: the question is the answer ?!(SPOLIER?)
Date: 13 Nov 92 16:01:14 GMT
Article-I.D.: husc3.1992Nov13.110115.17407
 <1992Nov12.184220.17456@CSD-NewsHost.Stanford.EDU> <1992Nov12.195756.16889@cbnewsl.cb.att.com>
Distribution: usa
Organization: Harvard Math Department

In article <1992Nov12.195756.16889@cbnewsl.cb.att.com>
piz@cbnewsl.cb.att.com (G G ZIPFEL) writes:
>>>>>>>>Q: How many letters are there in the answer?
>A: Exactly Ten.

As opposed to:

A: Oh, about a dozen. 

:-)

--Noam D. Elkies (elkies@zariski.harvard.edu)
  Dept. of Mathematics, Harvard University
===========================================================
fermat(p,v)= v=cf(p/abs(centerlift(sqrt(mod(-1,p)))));\
	[1,0]*pnqn(vector(length(v)\2,n,v[n]))
Newsgroups: rec.puzzles
From: more_da@pts.mot.com (DAN MORERA)
Subject: Re: the question is the answer ?!
Reply-To: more_da@pts.mot.com
Organization: Paging and Wireless Data Group
Date: Thu, 12 Nov 1992 19:53:21 GMT

In article 92Nov12084243@kbsw3.kbsw1.uucp, dougg@kbsw1.uucp (Doug Gerecht 806 1031) writes:
> >>>>> On 11 Nov 1992 14:44:30 -0500, berninge@cs.purdue.edu (James Berninger) said:
> 
> > NNTP-Posting-Host: alice.cs.purdue.edu
> > kanad@quip.eecs.umich.edu (Kanad Chakraborty) writes:
> > jpm10@cd.amdahl.com (Jim Millar) writes:
> 
> >>>dmuntz@quip.eecs.umich.edu (Daniel A Muntz) writes:
> >>>>thomas@sees.bangor.ac.uk (Thomas Varsamidis) writes:
> >
> >>>>>This is somewhat different, but the idea is the same:
> >>>>>
> >>>>>Q: How many letters are there in the answer?
> >>>>>A: Four
> >>>>A: 0
> >>>A: Eleven plus two
> >>A: Seventeen plus zero
> > A1: I
> > A2: II
> > A3: III
> > A4: Uh, seven?
> A: This one has eighteen
> -- 
> 
> 		dougg@kentek.com	Doug Gerecht


A^:  If I understand your question correctly, I would probably guess sixty-one.
---
*********************************************************************
* "Could be worse -- could be raining."  * Daniel Morera            *
*********************************************************************
*                                        * more_da@pts4.pts.mot.com *
*********************************************************************
From: J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger)
Newsgroups: rec.puzzles
Subject: Re: Word sequence
Date: 9 Nov 92 17:37:12 GMT
Organization: Dartmouth College, Hanover, NH
X-Posted-From: InterNews1.0a5@newshost.dartmouth.edu

I propose the following sequence of four-letter words:

STIR, TIRE, IRES, REST, ESTI, STIR, .... (forming an infinite
sequence!)

:-)

Since the puzzle doesn't specifically say all the words had to be
English, I deemed it acceptable to use the word "esti", which is the
Esperanto(!) word for "to be".

:-)

--Ted Schuerzinger
email: .zed@Dartmouth.EDU
"I should have realized it would be bad vodka when all the label said
was 'Russian Vodka'."
From: grue@cs.uq.oz.au (Pauli Dale)
Newsgroups: rec.puzzles
Subject: Re: Word sequence
Date: 13 Nov 92 06:44:54 GMT
Reply-To: grue@cs.uq.oz.au

Barry.John.Good@brunel.ac.uk (Barry John Good) writes:

>I hope this isn't an FAQ, but what's the longest sequence of 4-letter words 
>that you can find such that the last three letters of each word are the
>first three letters of the next in the sequence?


More results.......

7 letter words, length = 4:
	grasper raspers asperse sperses
	grasper raspers asperse spersed

8 letter words, length = 3:
	atropine tropines ropiness
	aventail ventaile entailed
	aventail ventaile entailer
	scrumple crumples rumpless
	sprattle prattler rattlers
	striddle triddler riddlers

9 letter words, length = 3:
	restamped estampede stampeded
	restamped estampede stampeder
	restamped estampede stampedes

For 10 & 11 letter words, the length is 2 and there are lots of them.



My current best for the original problem is 111:

ABAC bach ache chew hewe ewer weri eric rich icho chob hobo obol bola olam lama
amar mara arad rada adad dada adat data atap tapa apar pari aril rile ilea leas
east asta staw tawa awag wage agen gena enam name amel mela elan lana anan nana
anal nala alan lang anga ngai gaid aide idem demi emit mite iter tera eral rale
alen leno enol nole olea leal eale alec lech echo chok hoke oket keto eton tone
oner nere ered rede edea deas ease asem semi emir miro irok roka okas kasa asar
sari arid ride idea dead eadi adit dite item teme emes mese eses sess ESSE

And I have given up that search since it is unlikely to ever finish.



        						Pauli

Paul Dale                       | grue@cs.uq.oz.au
Department of Computer Science  | +61 7 365 2445
University of Queensland        |
Australia, 4072                 | Did you know that there are 41 two letter
                                |     words containing the letter 'a'?
--
Newsgroups: rec.puzzles
From: amorgan@Xenon.Stanford.EDU (Crunchy Frog)
Subject: Re: the question is the answer ?!
Organization: Computer Science Department, Stanford University.
            <1drnquINN8pr@alice.cs.purdue.edu> 
            <DOUGG.92Nov12084243@kbsw3.kbsw1.uucp>
Distribution: usa
Date: Thu, 12 Nov 1992 18:42:20 GMT

In article <DOUGG.92Nov12084243@kbsw3.kbsw1.uucp> 
  dougg@kbsw1.uucp (Doug Gerecht 806 1031) writes:
>>>>>> On 11 Nov 1992 14:44:30 -0500, 
  berninge@cs.purdue.edu (James Berninger) said:
>>>>>>
>>>>>>Q: How many letters are there in the answer?
>>>>>>A: Four
>A: This one has eighteen

A: This one has eighteen added to ten
A: This one has eighteen added to ten added to ten
A: etc

C "Not original" Frog
Newsgroups: rec.puzzles
From: sjdaskal@unix.amherst.edu (Steve)
Subject: Re: Great turtle race *Spoiler*ish
Organization: Amherst College, Amherst MA
Date: Fri, 13 Nov 1992 05:51:19 GMT

In article <1992Nov12.192101.13105@hellgate.utah.edu> tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>It was the day for the great turtle races, and Alfred, Bob, Carl, Dave,
>Ed and Frank shuffled up to the starting line.  There would be two
>races, and all of them were very excited (as excited as turtles can be!)
>They would average each turtles place to find their overall rank..
>
>In the first race:
>   Alfred came in before Bob or Carl did
>   Dave came in two places before Frank
>   Three turtles ran times in between Carl and Ed
>
>In the second race:
>   Alfred came in before Bob or Carl did, and ran better this race than before
>   Dave came in two places before Frank
>   Three turtles ran times in between Carl and Ed
>
>Quite peculiar!  Yet no turtle placed the same in both races.  Happily,
>after they computed their overall rank for the day every turtle got
>a medal.. gold, silver or bronze for 1,2 or 3rd.
>
>  What were the placings for each turtle in both races?  Who got the gold?

Well, perhaps I am wrong, but it seems to me that this is impossible.  Here
is my logic:  (please point out the flaw)

(I shall write simply the first letter of each turtle's name)
A came in before B and C in both races.  Thus, he came in at least fourth
both times.  Since he did better the second time, and can only win one medal
(come in 1-3), he must have com ein fourth the first time, meaning that B and
C were 5th and 6th, although not necessarily in that order.  (an argument
can be made to show that that order is correct, but it is irrelevant here.)
Therefore, D,E and F took medals in the first race, and must finish in the
bottom half of the second race.  (again, it can be shown that their order was
alphabetical, but that is irrelevant).
In the second race, D must finish 2 spots before F, indicating that D is 4th
and F is 6th.  Thus, E is 5th.
In addition, there are supposed to be three turtles between C and E, indicating
that C must be 1st.  (then the 2nd, 3rd, and 4th turtles are between).
However, A is supposed to do better than C, which is impossible if C is
1st.


Please show me where, if anywhere, this logic fails.
Thanks.

--Steve

Newsgroups: rec.puzzles
From: dagibbs@quantum.on.ca (David Gibbs)
Subject: Re: Another plurals puzzle
Date: Tue, 10 Nov 92 03:38:12 GMT
Organization: Quantum Software Systems

In article <1992Nov08.153848.24287@eng.umd.edu> rabino@eng.umd.edu (Michael Rabinovitz) writes:
>	What words, other than necropolis, have their plurals formed by
>dropping the final letter?

There are, of course, the obvious answers: metropolis and megapolis.
-David
(dagibbs@quantum.qnx.com)
From: rabino@eng.umd.edu (Michael Rabinovitz)
Newsgroups: rec.puzzles
Subject: Another plurals puzzle
Date: 8 Nov 92 15:38:48 GMT
Article-I.D.: eng.1992Nov08.153848.24287
Organization: College of Engineering, University of Maryland, College Park


	What words, other than necropolis, have their plurals formed by
dropping the final letter?

-- 
         	                        Mike Rabinovitz 
					rabino@eng.umd.edu
From: cab@col.hp.com (Chris Best)
Newsgroups: rec.puzzles
Subject: Re: Does anybody know what a TRYPTYCH is?
Date: 6 Nov 92 15:42:11 GMT
Article-I.D.: col.1992Nov6.154211.26930
Organization: your service

bickford@jimmy.dfci.harvard.edu (Paul Bickford) writes:
> A tryptych is a three paneled painting...

----------

No, No!  It's AAA's road map for dolts.
From: goddard@NeXTwork.Rose-Hulman.Edu (Bart E. Goddard)
Newsgroups: rec.puzzles
Subject: Re: gays and condoms
Date: 5 Nov 1992 13:53:43 GMT
Organization: Computer Science Department at Rose-Hulman
Reply-To: goddard@NeXTwork.Rose-Hulman.Edu (Bart E. Goddard)
NNTP-Posting-Host: g214-1.nextwork.rose-hulman.edu

In article <SUOPANKI.92Nov5001903@zombie.oulu.fi>  
suopanki@zombie.oulu.fi (Heikki Suopanki) writes:
>[stuff deleted]
> -Heikki

We have a few conventions among us posters, which include smileys:-),
and "SPOILER" and "LONG".  I would like to suggest that we institute 
another: "NFTEO" = not for the easily offended.

Bart
Xref: sparky rec.puzzles:7189 sci.math:14759
From: jgk@osc.COM (Joe Keane)
Newsgroups: rec.puzzles,sci.math
Subject: Re: math puzzles (answers)
Summary: Use a variant Euler-Maclaurin.
Keywords: secant
Date: 11 Nov 92 04:05:40 GMT
Reply-To: Joe Keane <jgk@osc.com>
Followup-To: rec.puzzles
Organization: Versant Object Technology, Menlo Park, CA
Weather: sunny, high 70, low 41
Moon-Phase: waning gibbous (99% of full)

In article <1992Nov4.142040.5774@infodev.cam.ac.uk> cet1@cus.cam.ac.uk (C.E.
Thompson) writes:
>1/(n+1) - 1/(n+3) + 1/(n+5) - 1/(n+7) + ...  ~
>
>            1/(2n) - 1/(2n^3) + 5/(2n^5) - 61/(2n^7) + 1385/(2n^9) - ...
>
>where 1, -1, 5, -61, 1385, ... are the Euler numbers. And this is indeed the 
>case, in the asymptotic sense (the series is not convergent, but the error is
>always less than the first term omited). 
>
>However, I wouldn't like to show you my current proof that the Euler-Maclaurin
>formula reduces to this simple form for this series, as it is quite horrible.

I'm amazed that you did it at all.

>Can anyone produce a concise and/or elegant proof?

I'll use operators to make things concise and elegant; D is the derivative
operator, and E is the shift operator.  In this notation, Taylor's theorem is:

E = exp D

You can do a bit of algebra and get the following:

(1 - E)^-1 = 1/2 (1 - coth (1/2 D))

This is the Euler-Maclaurin formula.  But it's not really what we want.  The
series should be alternating, and the sample points should be offset from the
derivatives.  We can try the following one:

E (1 + E^2)^-1 = 1/2 sech D

I'll expand this out in case you don't follow the operator stuff:

  inf                           inf
 ----                          ----   E
 \        j                  1 \       j    (2 j)
  >   (-1)  f(n + 2 j + 1) = -  >   ------ f     (n)
 /                           2 /    (2 j)!
 ----                          ---- 
 j = 0                         j = 0

(As usual, thanks to Emacs Calculator for the typesetting.)

The Euler numbers come from the secant series, just like the Bernoulli numbers
come from the tangent series in the Euler-Maclaurin formula.  I don't know if
this formula has a specific name; i just consider it a variant of the
Euler-Maclaurin formula.

You can play around with this formula, trying out different functions for f.
It usually works, even though both series often don't converge.  If f is a
polynomial, the series on the right terminates, so this is easy to test.  You
can even derive it this way, since the Euler numbers can be defined in terms
of differences of polynomials.

In the case we're talking about, f(n) is n^-1, so f^(j)(n) is j! n^-(j+1).
The factorials cancel and we're left with just the Euler numbers.  This gives
exactly the formula above for the error in the Leibniz series.

--
Joe Keane, amateur mathematician
jgk@osc.com (uunet!amdcad!osc!jgk)
From: dos105@rsphy2.anu.edu.au (Darren O'Shaughnessy)
Newsgroups: rec.puzzles
Subject: Re: Sums and products (was Re: Synonyms with different meanings)
Followup-To: rec.puzzles
Date: 6 Nov 1992 06:24:02 GMT
Organization: Australian National University
Distribution: world
NNTP-Posting-Host: 150.203.49.43

In article <1992Nov1.023450.20959@williams.edu>, 95csk_2@williams.edu
(Christopher S Kollett) wrote:
> However, I now have a new and improved answer which I hope satisfies
> all of the conditions in the question:
> 
> 1001 = 7 * 11 * 13 = 10^3 + 1^3.
> 

  It does. I was a bit disappointed to see that respondents used brute
force methods, like calculator/computer and a few minutes of programming or
trial and error. This problem interested me because knowing that a number
is the sum of two cubes leads immediately to the factorization:
    a^3 + b^3 = (a + b)(a^2 - ab + b^2)
which cuts down the search space considerably.

--
Signed,            |   "It is not Life itself which is precious, but
  Pat Rourke       |    Quality of Life. However, Quality of Life is
                   |      not an Objective thing; it is up to the
(you may not have  |        individual to decide what te values."
   heard of me)    |
Newsgroups: rec.puzzles
From: sillke@math25.mathematik.uni-bielefeld.de (Torsten Sillke)
Subject: dissecting the cube
Date: Wed, 11 Nov 1992 14:35:44 GMT
Organization: Math MadHouse Bielefeld, Germany

Martin Gardner [*] says, there is a unique dissection of the square into
five congruent parts.               ^^^^^^
                                    Proof?

[*] MG5SA: The Unexpected Hanging and Other Mathematical Diversions
    MG5SA: Simon & Schuster (1968)
    MG5SA.15.5 Dissecting a Square

Question:
For which k exists a dissection of the n-dimensional hypercube into
conguent connected parts which are not boxes?

If 1<k<=n, there is such a dissection. If in addition k is a prime, there
are infinitely many dissections.

Torsten Sillke
Newsgroups: rec.puzzles
From: thomas@sees.bangor.ac.uk (Thomas Varsamidis)
Subject: Re: How many letters in the answer?
Organization: University of Wales (Bangor)
Date: Fri, 13 Nov 1992 14:10:55 GMT

In article <1992Nov12.044032.389@dartvax.dartmouth.edu>, J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:
> In article <1drnquINN8pr@alice.cs.purdue.edu>
> berninge@cs.purdue.edu (James Berninger) writes:
> 
> > kanad@quip.eecs.umich.edu (Kanad Chakraborty) writes:
> > jpm10@cd.amdahl.com (Jim Millar) writes:
> > >>dmuntz@quip.eecs.umich.edu (Daniel A Muntz) writes:
> > >>>thomas@sees.bangor.ac.uk (Thomas Varsamidis) writes:
> > >>>>This is somewhat different, but the idea is the same:
> > >>>>
> > >>>>Q: How many letters are there in the answer?
> > >>>>A: Four
> > >>>A: 0
> > >>A: Eleven plus two
> > >A: Seventeen plus zero
> > A1: I
> > A2: II
> > A3: III
> > A4: Uh, seven?
> 1. Cinco [in Spanish]
> 2. Vier [in German]
> 3. Tri [in Russian]
> 4. Du [in Esperanto]
> 5. Five plus eight
> 6. Two times ten-and-one-half
> 7. Twenty minus five
> or, in Pig Latin:
> 8. Evensay.
> 

9. Pente [in Greek]

> --Ted Schuerzinger
> email: .zed@Dartmouth.EDU
> "I should have realized it would be bad vodka when all the label said
> was 'Russian Vodka'."

Thomas
From: Tim.Epstein@f526.n635.z3.fidonet.org (Tim Epstein)
Sender: acsnet@csource.oz.au
Newsgroups: rec.puzzles
Subject: Forumula for Pi
Date: Thu, 05 Nov 1992 23:05:04 +1000

Pardon me if this is slightly off topic, but with the "Pi" puzzle 
recently posted here I can't help asking. 
 
Could someone please provide me with a formula I can use to write a 
simple BASIC program to calculate Pi to any number of digits. Even 
better - If someone actually has a BASIC program lying around that 
does this, could they post it if it isn't too many lines, or send it 
to me (Eppy@yoyo.cc.monash.edu.au). 
 
I memorized Pi many years ago, but now only remember it to less than 
100 places and require the program so I can have another go. 
 
Many thanks,
Tim
 * Origin: I am serious ... and don't call me Shirley. (3:635/526)
Newsgroups: rec.puzzles