HDTenEightyP/NetNews huggingface dataset:
From: jtchern@ocf.berkeley.edu (Joseph Hernandez)
Newsgroups: rec.puzzles
Subject: Re: order of numbers?
Date: 7 Apr 1992 00:03:52 GMT
Organization: Joseph TeleCommunications Enterprises; Berkeley, California, USA
Distribution: JTC Enterprises Global Electronic Mailing Services
NNTP-Posting-Host: plague-ether.berkeley.edu

In article <1992Apr6.190547.27198@Princeton.EDU> 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
>Latin.
>The Romance languages are similar:
>5, 2, 8, 9, 4, 7, 6, 3, 1
>and
>5, 2, 9, 8, 4, 7, 6, 3, 1

WRONG! In Spanish, one of the Romance languages, the order would be different.

5, 4, 9, 8, 6, 7, 3, 1

(cinco, cuatro, nueve, ocho, seis, siete, tres, uno)

Zero (cero) would be before the 5 if it was to be used.


Just my 8% of a quarter--

Peter

-- 
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Newsgroups: rec.puzzles
From: mblum@chaos.cs.brandeis.edu (Cerebus)
Subject: Re: SPOILERS: Isaac Asimov's mysteries puzzle
Organization: Brandeis University
Date: Tue, 14 Apr 1992 04:53:16 GMT

russotto@eng.umd.edu (Matthew T. Russotto) writes:

[stuff deleted]

>No, you are right.   The joke is

>DEC 25 = OCT 31.
>(DEC= decimal (base 10), OCT = octal (base 12))
                                            ^^
					Don't you mean 8??!!
-- Matt
Matt "Matt" Blum  		mblum@chaos.cs.brandeis.edu

The city, 1:00am. People in their rooms, asleep, lonely, depressed.
One of them gets the urge to pet a small, furry animal.  That's where
I come in.  My name's Friday.  I carry a badger.
From: mdeboer@fwi.uva.nl (Marc de Boer (I88))
Newsgroups: rec.puzzles
Subject: C-puzzle (Tough)
Summary: Tricky C-programs; what do they do?
Keywords: puzzle
Date: 7 Apr 92 12:50:18 GMT
Organization: FWI, University of Amsterdam

Ok, smart dudes! Here are some mean, MEAN c programs!

Question: figure out what they do! And how do they do it?
Try to predict a few. Don't give up soon!

They should compile under unix (cc).
I haven't tried Turbo C, but it should be possible.

#include <stdio.h>
int Oo[1000],*O=Oo;main(o,O0)char**O0;{if(--o!=!0){*++O=atoi(*++O0);*++O= atoi(*++O0);for(O--;O!=Oo;){for(o=Oo-O;o;printf("F(%d,",O[++o]));printf( "%d",O[++o]);while(printf("%c",*(")\n"+(--o==Oo-O))),o!=Oo-O);(*O)?(*++O? ((*(O--))--,(*(O++))--,(*(O+!0)=*O),(*O=*(O-!!!0)+!0)):((*O)++,(*(--O))--)):((*O=*(O+!0)+!!!0),*(O --));}printf("%d\n",*++O);}}


main(o,O0)char**O0;{int OO,O;O=--o?atoi(O0[!0]):! 0;for(o=((OO=O*O)-O+!0+!0)>>!0;OO;o+=((--OO%O)?-! 0:((((o-!0)%O)?O:0)+!0))-(((o-!0)%O)?O:0)){printf ("\n%*d "+!!(OO%O),!0<<!0<<!0,o+=(o<!0)?O*O:0);}}

Good luck!


--------------------------------------------------------------------
Thanx for you input!                                   Marc de Boer
(/dev/null loved it!)                           (mdeboer@fwi.uva.nl)
--------------------------------------------------------------------

-- 
--------------------------------------------------------------------
Thanx for you input!                                   Marc de Boer
(/dev/null loved it!)                           (mdeboer@fwi.uva.nl)
--------------------------------------------------------------------
From: meng@idt.unit.no (Tore Mengshoel)
Newsgroups: rec.puzzles
Subject: Re: A simple sequence of numbers ...
Date: 10 Apr 92 08:40:16 GMT
Reply-To: meng@idt.unit.no (Tore Mengshoel)
Organization: Div. of CS & Telematics, Norwegian Institute of Technology

In article <mblum.702838289@chaos.cs.brandeis.edu>, mblum@chaos.cs.brandeis.edu (Cerebus) writes:
|> meng@idt.unit.no (Tore Mengshoel) writes:
|> 
|> >In article <1992Apr7.220710.27512@Cadence.COM>, hemmady@Cadence.COM (Shankar Hemmady) writes:
|> >|> 
|> >|> 
|> >|> Continue this sequence:
|> >|> 
|> >|> 	3, 3, 5, 4, 4, 3, 5, 5, 4, 3, ...
|> >|>
|> >e.g. 3, 5, 4, 4, 3, 3, 5, 5, 4, 3, .... 
|> 
|> What do you mean, "e.g. . . ." ??
|> The answer is:
|> 
|> 6, 6, 8, 8, 7, 7, 9, 9, 6, . . .
|> 
|> Isn't it??
|> 
|> -- Matt
|> Matt "Matt" Blum  		mblum@chaos.cs.brandeis.edu
|> 
Yes, that was the obvious answer in thread with numerous similar
sequences that has been presented here lately. By writing "e.g",
I implied that one can view the sequence to be something else
from the information given by the initial numbers. This interpretation
was: 3s, 5s, and 4s cyclically appearing in that sequence, each
appearing twice if it appeared once the last time and vice versa.

|> The city, 1:00am. People in their rooms, asleep, lonely, depressed.
|> One of them gets the urge to pet a small, furry animal.  That's where
|> I come in.  My name's Friday.  I carry a badger.

Tore
From: ed@wente.llnl.gov (Ed Suranyi)
Newsgroups: rec.puzzles
Subject: Re: Another Isaac Asimov puzzle -- geography
Date: 16 Apr 92 02:20:22 GMT
Organization: UC Davis Dept of Applied Science at LLNL

In article <MITCHELL.92Apr15182619@gluttony.uchicago.edu> mitchell@cs.uchicago.edu writes:
>This is the correct answer.
>This is one of my favorite puzzles of this sort.

That's absolutely right!  If anybody didn't see his solution, take a look
at it.

For the record, the first person to e-mail me with the correct solution
was Matthew P. Wiener.

Interestingly, everybody who's gotten it remembers it from Asimov's essay. 

Ed
ed@wente.llnl.gov
Newsgroups: rec.puzzles
From: farrell@coral.cs.jcu.edu.au (John Farrell)
Subject: Re: Dying young
Organization: James Cook University
Date:  7 Apr 92 02:29:54 GMT

In <7452@uqcspe.cs.uq.oz.au> warwick@cs.uq.oz.au (Warwick Allison) writes:
>tolman%asylum.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>>You are put in a room by evil Nazi's.

>1. Assume everyone else does whatever you do.
>2. Pick a number between 1 and 100 (using dice).  If the number is 1 or
>two, push the button.  

>I'm not very good at probability (I have two sisters, one of which is
>younger than the other), but there must be some assignment of probabilities
>that is optimal, since 100% and 0% are both fatal (all people die).

  Well, are we talking about optimal for you or optimal for everyone? The
simple fact is that you cannot lose if you don't press the button. You can
make a massive gain for everyone else, but so what if you're dead? Before you
can talk about an optimal strategy you have to define what optimal means.

  I am not stupid, I am not expendable, and I am not pressing the button.


Friendless




>Warwick.
>--
>  _-_|\      warwick@cs.uq.oz.au            / pinots pintos /
> /     * <-- Computer Science Department,  / piston pitons /
> \_.-._/     University of Queensland,    / points potins /
>      v      Brisbane, Australia.        / ------ ------ /
Newsgroups: rec.puzzles
Subject: Hidden Phrases   *** SPOILER ***
From: mulvenad@mof.govt.nz
Date: 8 Apr 92 13:58:00 +1200
Organization: New Zealand Ministry of Forestry

Here's the answers to the hidden phrases puzzles I posted last week
for those who need them.  If you don't want the answers don't read
the rest of this posting.





















1.  man overboard

2.  a backwards glance

3.  hole in one

4.  down town

5.  six of one and half a dozen of the other

6.  scrambled eggs

7.  turn back in anger

8.  rings (or circles) under the eyes

9.  three degrees below zero

10. neon light

11. reading between the lines

12. down in the dumps
-- 
****  Why is the symbol for anarchy always written the same way?  ****
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Newsgroups: rec.puzzles
Subject: Re: Is 11111111111 prime?
Date: 12 Apr 92 23:20:00 GMT
Organization: Texas A&M University, Academic Computing Services
News-Software: VAX/VMS VNEWS 1.41

>>> Prove or disprove that the number of primes in the sequence
>>> 1 , 11 , 111 ,  1111 , 11111 , ...
>>> is finite.    [DGC note:  There is only 1 prime: 11 .]

carabalo@phoenix.Princeton.EDU (David G. Caraballo) writes...
>Let a_n be the term with n 1's.  If n is even, then 11 divides a_n, so only
>11 is prime.  If n is odd, 9*a_n + 1 = 10^(n+1), so a_n = (10^(n+1) - 1)/9.
                                        ^^^^^^^^
>The numerator is a difference of 2 squares, and the rest is trivial:  if n>1
>is odd then a_n is composite.                  DGC, Dept. of Math, Princeton

Should that be 10^n, thus ruining the proof?

I will note that the 'even n' argument given above applies to all composite n.
Thus we are concerned with prime n. If someone thinks they have a solution,
they should try to factor a_11 = (10^11-1)/9.

Hint: it has two factors, the smaller one has 5 digits.

Dave Ring
dwr2560@zeus.tamu.edu
Newsgroups: rec.puzzles,soc.culture.indian
From: amorgan@Xenon.Stanford.EDU (Crunchy Frog)
Subject: Re: A simple number puzzle
Organization: Computer Science Department, Stanford University.
Date: Tue, 7 Apr 1992 23:29:42 GMT

In article <1992Apr7.202400.26466@Cadence.COM> 
  hemmady@Cadence.COM (Shankar Hemmady) writes:

>The puzzle goes as follows:
>
>"My child's age raised to itself gives my age.
>If my wife is younger than me by my child's age
>and if we are LIVE humans, then what are our ages?"
>
>a small clarification:  if my child's age is x,
>then my age is x**x (x to the power of x)
>And my wife is younger than me by x years.

Me: 27, Wife: 24, Brat: 3

The other solutions involve either child brides or women giving birth
at 60, which I rejected.

>- Shankar Hemmady
>hemmady@cadence.com

C Frog
Newsgroups: rec.puzzles
From: farrell@coral.cs.jcu.edu.au (John Farrell)
Subject: Re: Number puzzle (dollars per hour)
Organization: James Cook University
Date: 14 Apr 92 01:38:33 GMT

In <1992Apr13.154856.26408@cs.cornell.edu> karr@cs.cornell.edu (David Karr) writes:
>Let's hear from some of the "dying young" people on this one.

  OK then. I haven't seen my wife for two months as it is and I'm mighty
pissed off about it. One year is right out, I'm staying here!


Friendless
From: roger@quantime.co.uk (Roger Phillips)
Newsgroups: rec.puzzles
Subject: Re: 2 oldies  <*SPOILER*>
Keywords: water chestnut
Date: 13 Apr 92 00:44:43 GMT
Reply-To: roger@quantime.co.uk (Roger Phillips)
Organization: Quantime Ltd, London (software for market research)

In article <sjs.702662127@hoodlum>,
sjs@hood.hood.caltech.edu (Stephen Speicher) writes:
>2. Consider the letters H I J K L M N O
>   There is a one-word solution to this.

Water.
I've seen it used as a crossword clue,
and it was a chestnut even then.

-- 
Roger Phillips                            roger@quantime.co.uk
"Cards came next to cross-word puzzles in Lucia's index of inanities."
From: rose@fsu1.cc.fsu.edu (Kermit Rose)
Newsgroups: rec.puzzles
Subject: flipping over n objects k at a time
Date: 2 Apr 92 16:40:56 GMT
Article-I.D.: mailer.1992Apr2.114056.22329
Reply-To: rose@fsu1.cc.fsu.edu
Organization: Florida State University
News-Software: VAX/VMS VNEWS 1.3-4

A cursory look at this puzzle suggests that it is only possible to flip 
over n objects k at a time if k divides n, in which case the number of 
flips is n/k.

Kermit@FSU

This is the signature file of Kermit Rose.

Believing in dichotomies leads to hardening of your logic.

What determines the existence of Free Will?

Peace is the way to peace.  Hostility is the way to hostility.
From: karr@cs.cornell.edu (David Karr)
Newsgroups: rec.puzzles
Subject: Re: Dying young
Date: 15 Apr 92 15:56:57 GMT
Organization: Cornell Univ. CS Dept, Ithaca NY 14853

In article <1992Apr14.192328.495@hls.com> ravi@hls.com writes:
>So sitting tight and hoping that someone else presses the button
>apart from being irrational, does in no way mean you are not 
>stupid.

This obviously begs the following questions:  How do you define 
rational?  How do you define stupid?

Also, again, the less obvious:  What constitutes an optimal outcome?
(E.g. is it most lives saved, or just your own life saved.)

-- David Karr
From: cjmoran@unix2.tcd.ie (Ben Dover)
Newsgroups: rec.puzzles,soc.culture.indian
Subject: Re: A simple number puzzle
Date: 8 Apr 92 14:03:36 GMT
Organization: Trinity College, Dublin

In <1992Apr7.202400.26466@Cadence.COM> hemmady@Cadence.COM (Shankar Hemmady) writes:


>The puzzle goes as follows:

>"My child's age raised to itself gives my age.
>If my wife is younger than me by my child's age
>and if we are LIVE humans, then what are our ages?"

>a small clarification:  if my child's age is x,
>then my age is x**x (x to the power of x)
>And my wife is younger than me by x years.

IF MY CHILD IS 1 THEN I AM 1.UNTRUE
IF MY CHILD IS 2 THEN I AM FOUR.(TOO YOUNG TO HAVE CHILDREN)
MY CHILD IS 3.I AM 27.WIFE 24:CORRECT
IF MY CHILD WAS 4 I WOULD BE 256.NO GOOD.I'M DEAD.

HOWEVER MY CHILD COULD BE 2&11MONTHS OR 3 & 1 MONTH OR THREE AN 4 DAYS
OR ANY POSSIBLE COMBINATION OF AGES

y
m
Newsgroups: rec.puzzles
From: wyg9633@uxa.cso.uiuc.edu (Wooed Sun)
Subject: Re: Math Problem (Solution?)
Organization: University of Illinois at Urbana
Date: Wed, 8 Apr 1992 16:04:55 GMT

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

>	The problem is nine is not '9' but '11' in base 8. Does he/she switch bases in
>the middle of the problem?  Or am I just using the wrong base?  Either seems
>very possible.  The only reason I used base 8 is that I'm lazy and that is the
>only base on my calculator that works.  Better solutions?

>					Mike
The base is not necessarily refined to 8 for the number to be divisible
by 9. Let's say the base is x and assume x > 9. Then,
4444 = 4*(x**3+x**2+x+1) = 4*(x**4-1)/(x-1)
If you check with your calculator, you can see 4444 is divisible by 3
(therefore, 4444**2 is divisible by 9, and so is 4444**4444) when
x = 11,12,14,17,...

Wooed Sun.
-- 
email : w-yang@uiuc.edu (wyg9633@uxa.cso.uiuc.edu)
Newsgroups: rec.puzzles
From: ramkiss@bullet.ecf.toronto.edu (RAMKISSOON  SHERWYN)
Subject: Re: factorial puzzle
Organization: University of Toronto, Engineering Computing Facility
Date: Fri, 10 Apr 1992 16:39:30 GMT

In article <4781@umriscc.isc.umr.edu> ecarsten@next1.cs.umr.edu (Edward Carstens) writes:
>
>How many zeros are at the end of the number, 1000! ?
>
>(Of course there are 3 zeros at the end of 1000 but the question asks for 1000  
>factorial.)
>
>-Ed Carstens
>Creator of JugglePro

SPOILERS!

(hope the control-L worked)


I  remember this type of question from my high school days. To find the number
of zeros  at the end  of n!, simply successively divide n by powers of 5 and 
add up the  integer portion of the quotients

ie. for 1000!: # zeros= int(1000/5)+int(1000/25)+int(1000/125)+int(1000/625)
                      = 200+40+8+1=249 zeros

The reason of course is that 10=5*2 therefore the number of 10's in n!
equals the number of 5's since   2's are abundant.

Sherwyn
=======
From: jtchern@ocf.berkeley.edu (Joseph Hernandez)
Newsgroups: rec.puzzles
Subject: Re: A simple sequence of numbers ...
Date: 8 Apr 1992 17:22:03 GMT
Organization: Joseph TeleCommunications Enterprises; Berkeley, Califronia, USA
Distribution: JTC Enterprises Global Electronic Mailing Services
NNTP-Posting-Host: monsoon.berkeley.edu
Summary: ****FIRST SPOILER ALERT****
Keywords: ****SECOND SPOILER ALERT****

In article <1992Apr7.220710.27512@Cadence.COM> hemmady@Cadence.COM (Shankar Hemmady) writes:
>Continue this sequence:
>   3, 3, 5, 4, 4, 3, 5, 5, 4, 3, ...

			****FINAL SPOILER ALERT****









































Each number in the sequence is the number of letters in the English names of
the natural numbers beginning with: 1 (one-3 letters), 2 (two-3 letters), etc.

This is the continuation up to 20 (twenty-6 letters):

		6, 6, 8, 8, 7, 7, 9, 8, 6, ...


Just my 8% of a quarter's worth--
Peter

-- 
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    *    *    *                 One Telecommunications Complex Center         
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Newsgroups: rec.puzzles
From: mag@suna3.cs.uiuc.edu (Tom Magliery)
Subject: Re: How to cluster US States?
Organization: University of Illinois at Urbana-Champaign
Date: Fri, 17 Apr 1992 18:03:24 GMT

gt2807a@prism.gatech.EDU (Benjamin H. Cowan) writes:
>
>  What is the smallest number of clusters that one can group the US states
>into such that at least one state in the cluster touches every other state
>in that cluster.  For example, Florida, Georgia, and Alabama would make
>a valid cluster because Florida touches every other state in that cluster.
>The addition of Mississippi to that cluster would make it invalid because
>no one state touches every other state within the cluster.
>
>  So, what is the minimum number of clusters that the continental US can
>be grouped in to?  Also, what is the biggest possible cluster that can be
>made, according to number of states in that cluster?  Please *post* any and 
>all answers to rec.puzzles.

by a bit of graph theoretical mumbojumbo, there can't be any cluster
bigger than 4 states.  (the dual graph of a 5-state cluster would be
K_5, which is non-planar.)  but a few minutes of staring at a map
yields no 4-state clusters.  the only possibility (i think) would be
if there are some funny border rules about where exactly are NY's and
RI's borders; maybe they touch out there somewhere.  then NY, CT, MA,
and RI would form a cluster.  (a weaker possibility would be UT, CO,
AZ, and NM, but then some of the "touchings" are in single
points...kinda bogus, i'd say.)

if there are no 4-clusters, then 19 clusters are necessary:  one for
AK, one for HI, one for ME (and at most one other), and therefore 16
for the other 46 states.  personally, i have yet to do better than 20,
which can be done many ways, but i haven't even tried to prove 20 is
minimal.

a 20-cluster clustering:  {AK}, {HI}, {WA,OR,ID}, {CA,NV,AZ}, {CO,UT,WY},
{NM,TX,OK}, {KS,MO,NE}, {MT,ND,SD}, {MN,WI,IA}, {IL,IN}, {MI,OH}, 
{KY,TN}, {AK,LA,MS}, {FL,GA,AL}, {SC,NC}, {WV,VA,MD}, {DE,PA,NJ}, 
{NY,VT}, {CT,RI,MA}, {NH,ME}.

mag
-- 
Tom Magliery      | WARNING:  Do not attempt to hang from towel, or insert
mag@cs.uiuc.edu   | your head into the towel loop.  Failure to follow these
                  | simple instructions can be harmful or injurious.
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Dying young
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Wed, 8 Apr 1992 20:28:34 GMT

In article <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?

>There is a rational solution.

The strategy that would minimize the number of lives lost is for me NOT
to press the button.

I'm assuming that what the Nazis say is true, and there are 100 other
people with buttons; also that if some push buttons, the others will
survive.  (Otherwise there's no point in pushing the button anyway!)
I'm also assuming I know nothing about the other 100 so they can be
considered a random sample of the population.

Now given any person picked at random and put in this situation, there
is SOME probability p that through blind altruism, miscalculation,
suicidal tendencies, self-imposed drawing of straws, or what have you,
he or she will push the button.  Let's say that p = 0.05 (i.e. 1 in
20), for sake of argument.  Then the probability that we all die is
0.95 to the 100th power (the probability that none of the other 100
random people pushes the button), or about 0.0059; on the other hand,
with probability 0.9941 those who don't push (expected value = 1 (me)
+ 0.95 * 100 = 96; yes, this is fudging a bit but it's close enough)
survive, so the expected number of survivals is 0.9941 * 96 = 95.43
persons.  Compare this to the case in which I push the button, in
which we statistically expect 5 others also to do so, with expected
number of survivals = 95.00.

Now judging from the responses on this topic, I'd say that in fact
p > 0.05, in which case my strategy only gets better.

Note that this is not just a selfish strategy; it works just as well
if each of the 101 prisoners has a button wired to the next one, as
long as p is not too small.  My object is to save as many lives as
possible, not just my own.  (Also note that my guess about p is
important.  If p is substantially less than 0.05, say p = 0.04, the
analysis shows that pushing the button saves more lives.)

-- David Karr
Newsgroups: rec.puzzles
From: mblum@chaos.cs.brandeis.edu (Cerebus)
Subject: Re: A question of time
Organization: Brandeis University
Date: Tue, 14 Apr 1992 23:25:37 GMT

ncramer@bbn.com (Nichael Cramer) writes:

>Well, one solution comes to mind (assuming that we give a little latitude
>to the phrase "both clocks were correct"):

>In short, he left the top of the stairs at some time between 8:55 and 8:59.
>(For purposes of this example, assume it was 8:57.)

>However, the clock at the top of the stairs was digital, with seven-
>segement-display characters.  Morevoer, the upper-right segement in the
>hours digit was burned out.
>                  --            --
>                 |  |          |  
>I.e, rather than  -- , it read  -- ,
>                 |  |          |  |
>                  --            -- 
>making the professor think the time was 6:57.  Five minutes later, at the
>bottom of the stairs he read the second clock showing the (correct) time
>9:02 (that is 3hr 5min later according to his reckoning).
                
Umm, 9:02 is TWO hours and 5 minutes after 6:57, not three.  Now, if
the lower left LED was *also* burned out, he would've thought it read
5:57 instead of 8:57, because it would've looked like this:
	--
       |
	--
	  |
	-- 

That's the only possible way he could've thought it was 3 hours and 5
minutes later, if the LED burnout answer is correct.  It would of
course be unlikely for those two LEDs in particular to burnout at the
same time, but nobody ever said these puzzles were supposed to be likely.

I don't like the solution much, but it's the only one I can think of.

-- Matt
Matt "Matt" Blum  		mblum@chaos.cs.brandeis.edu

The city, 1:00am. People in their rooms, asleep, lonely, depressed.
One of them gets the urge to pet a small, furry animal.  That's where
I come in.  My name's Friday.  I carry a badger.

Newsgroups: rec.puzzles