HDTenEightyP/NetNews huggingface dataset:
From: bdsm@fermat.WPI.EDU (Billy Don McConnell)
Newsgroups: rec.puzzles
Subject: Cute Problem  [SOLUTION]
Date: 14 Dec 1992 04:32:34 GMT
Organization: Worcester Polytechnic Institute
Distribution: world
NNTP-Posting-Host: fermat.wpi.edu
Originator: bdsm@fermat.WPI.EDU


This is a proposed solution to the "Cute Problem" proposed in article
<Bz7Go6.16F@world.std.com> by Jeff Kenton (jkenton@world.std.com):

|> Here's a problem I haven't seen before that I thought folks
|> might enjoy:
|> 
|> 	Given the numbers 101, 10101, 1010101, ... (decimal, not binary)
|> with 'n' 1's separated by 'n-1' 0's.  101 is prime.  Prove that none of
|> the rest is prime.

The solution follows...so, if you don't want to see the solution...
                        DON"T READ ANY FURTHER

Let's let  N(1) = 101
           N(2) = 10101
           N(3) = 1010101
               et cetera....

First, simply note that each N(n) is the sum of a Geometric Series
               
                                  1        2              n
                N(n) =   1  +  100   +  100   + ... +  100    ( n >= 1 )

and is therefore rewritable using the standard formula, which yields

              n+1                           n+1             2(n+1)
         a*( r    - 1 )   a = 1      1*( 100     - 1)   ( 10       - 1)
 N(n) =  --------------            = ---------------- = ---------------
            ( r - 1 )     r = 100       ( 100 - 1 )             99

                 2    
          (  n+1)             n+1           n+1 
        ( (10   ) -  1)   ( 10    + 1 ) ( 10    - 1 )
      = --------------- = ---------------------------
              99                      99

            n+1                                    n+1
        ( 10    + 1 ) ( string of n+1 '9's )   ( 10    + 1 )( string of n+1 '1's )
      = ------------------------------------ = -----------------------------------
                 99                                            11

Now, if  n  is odd, then 11 divides evenly into the second Factor of the
Numerator, making N(n) Composite...unless, of course, that second Factor is
exactly  11, which occurs when and only when  n = 1 (this case we know is Prime).

If n is even, then 11 divides evenly into the first Factor of the Numerator;  if 
you don't believe me, observe...

                    n+1                n+1                n+1
                ( 10    + 1 )   ( -(10)    - 1 )   ( (-10)    - 1 )
                ------------- = ---------------- = ----------------
                    10 + 1          -10 - 1           (-10) - 1

                           1        2              n
               =  1 + (-10)  + (-10)  + ... + (-10)  

                             2           n
               =  1 - 10 + 10  - ... + 10  = Positive Integer

Thus, in this case N(n) is Composite without exception.

****************************OED: Oper Edei Deixai******************************

Don Mc Connell
bdsm@poincare.wpi.edu
Newsgroups: rec.puzzles,sci.math
From: kanad@quip.eecs.umich.edu (Kanad Chakraborty)
Subject: Re: Circular graph problem
Organization: University of Michigan EECS Dept., Ann Arbor
Date: Wed, 16 Dec 1992 19:57:25 GMT

In article <1992Dec16.172145.21848@dartvax.dartmouth.edu> J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:
>In article <1992Dec16.113500.1157@client24.comlab.ox.ac.uk>
>mnewman@comlab.ox.ac.uk (Matthew Newman) writes:
>
>> Consider circular graphs with N nodes, such that there is an integer
>> at each node. Consider those integers K such that for some graph G
>> each integer in [1..k] is the sum of integers at the nodes of some
>> connected sub-graph G. For example with N=4 I can get K=11 with the
>> following graph :     (1)--(2)
>>                        |    | 
>>                       (3)--(5) 
>> 
>> and I can do better than this and get the maximum possible k=13 like
>> this :     (1)--(2)    and like this:   (1)--(3)
>>             |    |                       |    |
>>            (4)--(6)                     (2)--(7)
>
>I don't understand why these last two count as valid -- in the first,
>there's no combination that totals 4, while in the second there's no

yes, there is : 3+1 = 4

>combination that totals 7.  Or am I just reading yet another problem

there is a combination that totals 7, namely 4 + 1 + 2.

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


Kanad
Newsgroups: rec.puzzles
From: bhayes@Xenon.Stanford.EDU (Barry Hayes)
Subject: Re: Monty Hall Puzzle RIP
Organization: Computer Science Department, Stanford University.
Date: Fri, 18 Dec 1992 19:26:40 GMT

In article <92353.100332SHERE@SLACVM.SLAC.STANFORD.EDU> <SHERE@SLACVM.SLAC.STANFORD.EDU> writes:
>Alright, I've had just about all bickering I can stand on this
>Monty Hall drivel and I'm going to put it to bed.

Lee-

I am going to extend to you a wonderous offer.  We will set up the
test.  We will have three boxes, you will choose one.  I will put $200
of my own money in one of the boxes and nothing in each of the others.
After you have chosen a box, I will show you one of the other boxes as
empty.  You say it makes no difference if you switch or not, and so I
will charge you $100 to switch.  If it makes no difference, you'd
be a fool to pay to switch.

You now have, by your calculations, a fifty-fifty chance of having the
box containing $200, and so on average you would expect that I will
end up giving you $100 for each game.  If, at the start of each game,
you give me $90, you would expect to make $10 each time we play this
game.

I'm really offering to do this -- let's say 20 rounds of the game.  If
you believe your logic, I'm offering you $200.  

Wanna?
From: ketil@ii.uib.no (Ketil M. Malde)
Newsgroups: rec.puzzles
Subject: Re: Christmas riddle
Date: 11 Dec 92 17:31:01 GMT
	<1992Dec10.180613.6676@iscnvx.lmsc.lockheed.com>
Reply-To: ketil@ii.uib.no
Organization: student of numerical analysis at the University of Bergen, Norway

>>>>> On Thu, 10 Dec 92 18:06:13 GMT, record@force.ssd.lmsc.lockheed.com said:

In article <1992Dec9.083159@gruc19.nor.chevron.com>, hjiwa@gruc19.nor.chevron.com (Santa Claus Ho Ho Ho) writes:
|> What's the perfect Christmas gift for the man who has everything ???
|> What's the perfect Christmas gift for the woman who has everything ???
|> 
|> HO, HO, HO

In article <3305@nlsun1.oracle.nl>, asymes@nl.oracle.com (Andy Symes) writes:

>> A box to put it in!  Oh, maybe not.  I guess the man already has that.

Well, but does he have a box the box fits in?  And if so, does he have
an even larger one for that one?  ... etc. By induction, he has an
infinite amount of boxes.

Since we assume that the amount of boxes in the universe is finite,
this person doesn't exist, and so the what-for-nnn-for-christmas-
problem is solvable.  The converse also holds, of course.

In other words, if you there exists a person to whom it is impossible
to buy a christmas present, the amount of boxes in the universe must
be infinite.  No?

--
 = Ketil Malde                     In real life:  ketil@ii.uib.no =
 = Nuke the whales!                    Honk if you love unicorns! =
From: ST102315@brownvm.brown.edu (Jake)
Newsgroups: rec.puzzles
Subject: Re: Big Numbers
Date: Thu, 17 Dec 1992 02:15:37 EST
Organization: Brown University
Distribution: world
NNTP-Posting-Host: brownvm.brown.edu
News-Software: BNN via BNN_POST v1.0 beta

>>however, I just realized that only finitely many things can be described
>>by 150 characters or less.
>
>Disproof of that assertion by demonstration:
>        The positive integers
>        The integers
>        The real numbers
>        The complex numbers
>        The even integers
>        The odd integers
>Now, if you meant that there are only a finite number of descriptions
requiring
>150 or fewer characters, I'd have to agree with you.

Nitpicks, nitpicks, nitpicks...There are a finite number of descriptions
requiring 150 or fewer characters.  If we look at the set of all
descriptions with 150 or fewer characters that specify a unique positive
integer, we get a subset of the set whose members are described above,
and hence this subset has a finite number of elements.  (It's not empty
because, for instance, 1="one" can be described using 150 or less
characters.)  However, it was proved that all positive integers can be
described in 150 or fewer characters.  Hence, the number of elements in
the subset described above is greater than or equal to the number of
positive integers (greater than because "one" and "two minus one" are
both members of the subset).  Hence, the number of elements in the
positive integers is finite.  QED.

If anyone needs convincing that the number of descriptions requiring
150 or fewer characters (I prefer only allowing descriptions that
consist of letters and only letters), I can provide a proof, but I
don't feel like doing it now.  <tired smiley>

If movement were possible, Zeno would be turning over in his grave,
jwgh
Newsgroups: rec.puzzles
From: seeta@eng.wayne.edu (Seetamraju UdayaBhaskar Sarma)
Subject: Re: Typing teaser
Reply-To: seeta@eng.wayne.edu
Organization: College of Engineering, Wayne State University, Detroit Michigan, USA
Date: Tue, 15 Dec 1992 21:08:07 GMT

In article 29375@dartvax.dartmouth.edu, J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:
> A lot of times when I type I hit the wrong key.  I'll either hit one of
> the neighboring keys on the same row, or I'll hit one of the
> neighboring keys the same finger would hit.  Most of the time, I end up
> with nonsense words that the spell check discovers, but I began
> wondering:
> 
> What's the longest possible word where you can have such typos and end
> up with a different word?  What if you only make one typo per word?
> 
> Remember, I'm not talking about switching just any letter.  It has to
> be one of the key adjacent to the letter you're trying to type. (eg.
> for 'g' the four 'typos' would be f, t, h, and b.)
> 
> So far the longest word I've gotten is 'fathers' for 'gathers'.  But
> I'm sure the net can do much better.
> 
> --Ted Schuerzinger
> email: .zed@Dartmouth.EDU
> "I should have known it would be bad vodka when all the label said was
> 'Russian Vodka'."


An interesting seed for an algorithm that does spelling CORRECTION...



Seetamraju Udaya Bhaskar Sarma
(email : seetam @ ece7 . eng . wayne . edu)
From: libes@cme.nist.gov (Don Libes)
Newsgroups: rec.puzzles
Subject: Rubik's Royal Brain Teaser
Date: 17 Dec 92 05:27:45 GMT
Organization: National Institute of Standards and Technology

Two of us put the puzzle together in a couple of hours.  But the
teaser on the box implies that we were supposed to solve it in a more
deductive way.  Two questions that we never figured out are:

How can you predict the orientation of the corner jewels?
Do the white jewels fall randomly and do they introduce randomness
or can their appearance and behavior be predicted?

Don Libes          libes@cme.nist.gov      ...!uunet!cme-durer!libes
From: murray@rocky.raleigh.ibm.com (Charlie Murray)
Subject: Re: Circular graph problem
Date: Wed, 16 Dec 1992 20:58:02 GMT
Organization: IBM, Research Triangle Park, North Carolina

In article <1992Dec16.172145.21848@dartvax.dartmouth.edu>, J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:
|> In article <1992Dec16.113500.1157@client24.comlab.ox.ac.uk>
|> mnewman@comlab.ox.ac.uk (Matthew Newman) writes:
|> 
|> > Consider circular graphs with N nodes, such that there is an integer
|> > at each node. Consider those integers K such that for some graph G
|> > each integer in [1..k] is the sum of integers at the nodes of some
|> > connected sub-graph G. For example with N=4 I can get K=11 with the
|> > following graph :     (1)--(2)
|> >                        |    | 
|> >                       (3)--(5) 
|> > 
|> > and I can do better than this and get the maximum possible k=13 like
|> > this :     (1)--(2)    and like this:   (1)--(3)
|> >             |    |                       |    |
|> >            (4)--(6)                     (2)--(7)
|> 
|> I don't understand why these last two count as valid -- in the first,
|> there's no combination that totals 4, while in the second there's no
|> combination that totals 7.  Or am I just reading yet another problem
|> incorrectly?!?

A connected sub-graph can be one node.  In the first, the connected sub-graph
that totals for is the single node (4).  Likewise for (7) in the second.
However, in the second, there is no solution for 5.

--------------------------------------------------------------------
  Charlie Murray           CompuServe:  73707,1752
  Consulting at IBM        Internet:    73707.1752@compuserve.com

  NOTE:  If sending email use the addresses given in this .sig,
         not the one in the header of this message.
--------------------------------------------------------------------
Newsgroups: rec.puzzles
From: ttw@beta.lanl.gov (Tony Warnock)
Subject: Re: Ask Marilyn
Organization: Los Alamos National Laboratory
 
 
      Marilyn was right. This is an old puzzle. I used to have trouble
      explaining it to my students. A simulation is rather convincing.
 
 
                                      Tony Warnock
 
 
                  It signifies nothing to play well if you lose.
                                                    - Anonymous
Date: Mon, 14 Dec 1992 20:34:26 GMT
From: clong@romulus.rutgers.edu (Chris Long)
Newsgroups: rec.puzzles
Subject: Re: Ask Marilyn
Date: 14 Dec 92 21:07:13 GMT
Organization: Rutgers Univ., New Brunswick, N.J.

In article <92349.120251HERSCH@auvm.american.edu>, Herschel Browne writes:

> Does everyone agree that Marilyn vos Savant's answer to the pancake
> probability problem in yesterday's Parade magazine was wrong? The
> answer is really 50-50. Either you've picked the pancake with 2 brown
> sides, in which case the other side is brown, or you've picked the
> pancake with one brown side, in which case the other side is not
> brown. 50-50.

What's the question?  I wouldn't be so quick to say she's wrong,
since she seems to understand basic probability; are you sure events
in the sample space are equally likely?
-- 
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618

"S.B., with an I.Q. of 161, failed to complete his course of study,
running away instead with his professor's wife."
H. J. Eysenck, _Know Your Own I.Q._
From: wsbujb1@rwc.urc.tue.nl (Jan Stout)
Newsgroups: rec.puzzles
Subject: Re: Christmas riddle
Date: 15 Dec 92 14:57:01 GMT
Reply-To: wsbujb1@urc.tue.nl

asymes@nl.oracle.com (Andy Symes) writes:

>In article <1992Dec9.083159@gruc19.nor.chevron.com>, hjiwa@gruc19.nor.chevron.com (Santa Claus Ho Ho Ho) writes:
>|> What's the perfect Christmas gift for the man who has everything ???
>|> What's the perfect Christmas gift for the woman who has everything ???
>|> 
>|> HO, HO, HO
>Thats easy. In either case, a burglar alarm will do nicely :)
>Andy.
That's stupid, if they have got everything, they also have a burglar alarm.
The only thing they don't have is nothing!

Ivana
Newsgroups: rec.puzzles
From: blum@medici.ils.nwu.edu (Dan Blum)
Subject: Re: Ask Marilyn
Reply-To: blum@ils.nwu.edu
Organization: The Institute for the Learning Sciences
X-Newsreader: TIN [version 1.1 PL8]
Date: Wed, 16 Dec 1992 17:33:06 GMT

HERSCH@auvm.american.edu wrote:
> Does everyone agree that Marilyn vos Savant's answer to the pancake
> probability problem in yesterday's Parade magazine was wrong? The
> answer is really 50-50. Either you've picked the pancake with 2 brown
> sides, in which case the other side is brown, or you've picked the
> pancake with one brown side, in which case the other side is not
> brown. 50-50.

> H.

> Herschel Browne
> "The" American University

Well, I didn't see it, but if it's the problem where you have two
pancakes, one brown on both sides, the other white on one and brown on
the other, you put both in a bag, pull one out, look at one side, see
that it's brown, and want to know the probability that the other side
is brown, and she gave the answer "2/3", then she's right.  There are
three brown sides that one might be looking at with equal probability,
and in two of those cases the other side of the pancake is brown.
This is an old chestnut, and should be in the FAQL if it isn't already.
--
________________________________________________________________________
Dan Blum           Institute for the Learning Sciences   Room 327
blum@ils.nwu.edu   1890 Maple Ave., Evanston, IL 60201   708-467-2306

"I fingered the butt of my .45, fighting the urge to pistol-whip Plato."
________________________________________________________________________
Newsgroups: rec.puzzles
From: kanad@quip.eecs.umich.edu (Kanad Chakraborty)
Subject: Re: Cute problem radicalized
Organization: University of Michigan EECS Dept., Ann Arbor
Date: Tue, 15 Dec 1992 19:30:12 GMT

In article <1992Dec15.163443.29090@dartvax.dartmouth.edu> J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:
>In article <klausf.724434532@hathi>
>klausf@hathi.informatik.rwth-aachen.de (Klaus Frank) writes:
>
>> And here's another problem for those who are fascinated by 101 and 91: 
>> 
>> 	Given x, y such that the following equations hold for a number n:
>> 		0 <= n < 10^8
>> 		x = (n^13) mod (10^12 - 1)
>> 		y = (n^13) mod 10
>> 
>> 	Calculate n, but don't use just brute force.
>
>I may be getting the mod function mixed up (I haven't studied any math
>in two years!), but how about the following solution:
>
>n=x=y=1
>
>It seems to me that there are an infinite number of solutions -- just
>plug in various vales of n.
>
>OK, assuming integral values of n there isn't an infinite number of
>solutions, because n is bounded, but you get my drift.
>
>
>--Ted Schuerzinger
>email: .zed@Dartmouth.EDU
>"I should have known it would be bad vodka when all the label said was
>'Russian Vodka'."

Another solution :  n = x = y = 0.

Kanad
From: dik@cwi.nl (Dik T. Winter)
Newsgroups: rec.puzzles
Subject: Re: Bit Counting
Date: 13 Dec 92 12:43:54 GMT
Organization: CWI, Amsterdam

In article <4f9uyKu00VolEZ_Uk6@andrew.cmu.edu> court+@CMU.EDU (Court Demas) writes:
 > Excerpts from netnews.rec.puzzles: 9-Dec-92 Re: Bit Counting Warwick
 > Allison@cs.uq.oz (2099)
 > 
 > > The arithmetic Add(x,y) operation is MUCH more complex than
 > > the BitCount(x) operation.
 > 
 > Fine.  The problem called for O(lg n) *arithmetic* operations.  
 > 
Fair enough.  If I remember the algorithm well it was something like:
split the number in two numbers, one with the bits cleared in even
positions, one with the bits cleared in odd positions (counting from 0
for the LSB).  Shift to the right the number with bits cleared in odd
positions and add those two numbers.  Continue with clearing bits in
groups of 2, 4, 8 etc.  Was that the algorithm?  In that case of the
operations involved only *one* is arithmetic: the addition.  All others
are not more, nor less, arithmetic than the bitcounting itself.  So if
that is your definition of arithmetic operations the problem can be
solved in O(1) arithmetic operations.  Otherwise you had better inform
us what your definition of arithmetic operations is.  (I would say:
addition, subtraction, multiplication, division, exponentiation,
root finding, etc., etc., but not clearing specific bits, that is a
logical operation.)

If this was not the algorithm please send me the correct algorithm as it
has expired from our news directory.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl
From: elmo@cybernet.cse.fau.edu (Yuri Villanueva)
Newsgroups: rec.puzzles
Subject: Re: Ask Marilyn
Date: 15 Dec 92 20:39:52 GMT
Organization: Cybernet BBS, Boca Raton, Florida

HERSCH@auvm.american.edu writes:

> Does everyone agree that Marilyn vos Savant's answer to the pancake
> probability problem in yesterday's Parade magazine was wrong? The
> answer is really 50-50. Either you've picked the pancake with 2 brown
> sides, in which case the other side is brown, or you've picked the
> pancake with one brown side, in which case the other side is not
> brown. 50-50.
> 
> H.
> 
> Herschel Browne
> "The" American University

I think she got it right.  You're looking at a brown side, so...

The probability you picked the brown/brown pancake is 2/3
The probability you picked the brown/gold  pancake is 1/3
The probability you picked the  gold/gold  pancake is 0

.. given that you are looking at a brown side, i.e/, P(bb|b)=2/3,
P(bg|b)=1/3, and P(gg|b)=0.  And that's it.  The probability then that
the other side is also brown is simply the probability of you picking
the brown/brown pancake.  In other words, the answer is P(bb|b), 2/3.
Xref: sparky rec.puzzles:7987 sci.math:17072
From: J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger)
Newsgroups: rec.puzzles,sci.math
Subject: Re: Circular graph problem
Date: 17 Dec 92 00:33:42 GMT
 <1992Dec16.172145.21848@dartvax.dartmouth.edu>  
 <1992Dec16.195725.11131@zip.eecs.umich.edu>
Organization: Dartmouth College, Hanover, NH
X-Posted-From: InterNews1.0b1@newshost.dartmouth.edu

In article <1992Dec16.195725.11131@zip.eecs.umich.edu>
kanad@quip.eecs.umich.edu (Kanad Chakraborty) writes:

> In article <1992Dec16.172145.21848@dartvax.dartmouth.edu> J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:
> >In article <1992Dec16.113500.1157@client24.comlab.ox.ac.uk>
> >mnewman@comlab.ox.ac.uk (Matthew Newman) writes:
> >
> >> Consider circular graphs with N nodes, such that there is an integer
> >> at each node. Consider those integers K such that for some graph G
> >> each integer in [1..k] is the sum of integers at the nodes of some
> >> connected sub-graph G. For example with N=4 I can get K=11 with the
> >> following graph :     (1)--(2)
> >>                        |    | 
> >>                       (3)--(5) 
> >> 
> >> and I can do better than this and get the maximum possible k=13 like
> >> this :     (1)--(2)    and like this:   (1)--(3)
> >>             |    |                       |    |
> >>            (4)--(6)                     (2)--(7)
> >
> >I don't understand why these last two count as valid -- in the first,
> >there's no combination that totals 4, while in the second there's no
> 
> yes, there is : 3+1 = 4

But there's no 3 in the first graph -- only 1, 2, 4, and 6
> 
> >combination that totals 7.  Or am I just reading yet another problem
> 
> there is a combination that totals 7, namely 4 + 1 + 2.

And here there's no 4 -- just 1, 2, 3, and 7.


--Ted Schuerzinger
email: .zed@Dartmouth.EDU
"I should have known it would be bad vodka when all the label said was
'Russian Vodka'."
From: dik@cwi.nl (Dik T. Winter)
Newsgroups: rec.puzzles
Subject: Re: Supermarket puzzle
Date: 12 Dec 92 00:23:12 GMT
Distribution: rec
Organization: CWI, Amsterdam

In article <1992Dec8.230516.24387@informix.com> robertw@informix.com (Robert Weinberg) writes:
 > Max weight: bottled water - not the "name-brand" bottles
 > on the shelves, but the $.25/gallon "filtered" water that comes out
 > of the dispensers. That will give you about 3200 pounds for $100.
 > 
 > Volume: those giant bags of pre-popped popcorn, preferably day-old and
 > discounted.

Interestingly: my supermarket has neither.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl
Xref: sparky rec.puzzles:7905 rec.org.mensa:8789