HDTenEightyP/NetNews huggingface dataset:
From: dseal@armltd.co.uk (David Seal)
Newsgroups: rec.puzzles
Subject: Re: Missing subject header
Date: 13 Apr 92 14:07:12 GMT
Distribution: rec
Organization: A.R.M. Ltd, Swaffham Bulbeck, Cambs, UK

In article <1992Apr12.112505.12875@Princeton.EDU>
carabalo@phoenix.Princeton.EDU (David G. Caraballo) writes:

>>In article <1992Apr09.180554.4791@watson.ibm.com>, news@watson.ibm.com (NNTP News Poster) writes:
>>> 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 .]
>
>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

No. 9*a_n + 1 = 10^n, so a_n = (10^n - 1)/9, and the numerator is not the
difference of two squares.

As far as I know, the question of how many of these numbers are primes is
unresolved. They're known as repunit primes, by the way.

David Seal
dseal@armltd.co.uk

All opinions are mine only...
Newsgroups: rec.puzzles
From: macster@IASTATE.EDU (Michael C Mccarty)
Subject: The Philosopher's Puzzle... 
Reply-To: macster@IASTATE.EDU (Michael C Mccarty)
Organization: Iowa State University
Date: Thu, 9 Apr 1992 17:20:34 GMT

	Now that I've calmed down a bit about 'Dying Young' puzzle, I've just
remembered another problem that I came across a while back that is sort of
related to that puzzle.  I haven't been reading this newsgroup for very long so
I hope this one hasn't been done before. 

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

	I must note that this isn't really much of a puzzle but it is a good refresher
course in good, clear, logical thinking (which I think some of you are in dire
need of).

				Logically yours,
				Mike

I have no .sig (sigh).
Newsgroups: rec.puzzles
From: brent-b@eastern.COM (Brent McNamee)
Subject: Math (?) puzzle
Organization: Eastern's Public Access BBS, Toronto CAN
Date: Wed, 15 Apr 1992 16:12:13 GMT

Hey all...
 
You may have seen this one before, but here goes anyway...

What are the next two (or three, etc...) numbers in the following series?

NOTE: it is NOT a mathematical series...

1
11
21
1211
111221
312211
13112221
1113213211

...and so on?
From: BANDA@AUSVM1.VNET.IBM.COM ("Venu Prabhakar Banda")
Newsgroups: rec.puzzles
Subject: Math Problems II: Confidence Boosters !!
Date: 10 Apr 92 18:11:14 GMT

Here are some real quickies from high school days

1. Evaluate the following

      log(tan 1) + log(tan 2) + .... + log(tan 90) = ?

2. Replace the "+" in the expression above by "*" and re-evaluate

3. Solve for x and y in the following

      tan(x) + cot(x) = 2sin(y)

4. Solve for x, y in the natural number domain

      1! + 2! + 3! + ... + y! = x^2

   determine all possible solutions to the above

5. Show that the number 1111....(91 times) is composite

OK, Ok ...So they are all very easy and it took you all less than
5 seconds to crack all of them !!!.....
that they were asked in high school they are not bad.....

Anyway watch this space for "Math Problems-III" for tougher ones.

Cheers,

Venu
====
From: warwick@cs.uq.oz.au (Warwick Allison)
Newsgroups: rec.puzzles
Subject: Re: Dying young (solved)
Date: 7 Apr 92 06:17:52 GMT
Reply-To: warwick@cs.uq.oz.au

>Are we talking about optimal for you or optimal for everyone?
>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 define optimal to mean most likely to ensure your own survival...


Assuming there were two types of people in the world.  Js (who will
not ever press the button) and Ws (who roll dice to decide).
Let's see what various concentrations of Js and Ws do.  Pardoning the pun.

I used simulation to get these figures.

If Ws press the button 1% of the time:

   Before    After
   J    W    J    W
 100    0    0    0	ie. all Js - none press button -> all die
  90   10    9    0	ie. mainly Js -> usually all die
  80   20   13    3
  70   30   19    8
  60   40   17   11
  50   50   19   19
  40   60   17   25
  30   70   15   34
  20   80    9   37
  10   90    6   54
   0  100    0   57	ie. all Ws -> half the time, noone presses button
TOTAL     12730 25156   (over 100 random trials)


Now, if Ws press the button 2% of the time:

 100    0    0    0
  90   10   17    1
  80   20   29    6
  70   30   30   12
  60   40   36   23
  50   50   26   25
  40   60   27   40
  30   70   23   52
  20   80   16   63
  10   90    8   73
   0  100    0   82
TOTAL     21410 38139


Now 3%:

 100    0    0    0
  90   10   23    2
  80   20   29    6
  70   30   40   16
  60   40   46   29
  50   50   34   32
  40   60   30   44
  30   70   25   56
  20   80   17   68
  10   90    9   81
   0  100    0   92
TOTAL     25750 43252

4%:

 100    0    0    0
  90   10   26    2
  80   20   43   10
  70   30   46   18
  60   40   45   28
  50   50   45   42
  40   60   38   54
  30   70   29   64
  20   80   19   74
  10   90    9   85
   0  100    0   94
TOTAL     30360 47724
 
50%:

 100    0    0    0
  90   10   90    5
  80   20   80    9
  70   30   70   15
  60   40   60   20
  50   50   50   25
  40   60   40   30
  30   70   30   35
  20   80   20   39
  10   90   10   45
   0  100    0   51
TOTAL     45000 28001



The optimal depends on the distribution of Js and Ws in the population,
but for an even mix, about 7 or 8% seems best (but in an even mix, you are
better off being a J [see above tables at 50 50]!!!!!).

One thing is clear.  The more Ws there are in the world, the better :-)



Warwick
--
  _-_|\      warwick@cs.uq.oz.au            / pinots pintos /
 /     * <-- Computer Science Department,  / piston pitons /
 \_.-._/     University of Queensland,    / points potins /
      v      Brisbane, Australia.        / ------ ------ /
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Newsgroups: rec.puzzles
Subject: Re: Dying young
Date: 9 Apr 92 05:27:00 GMT
Organization: Texas A&M University, Academic Computing Services
News-Software: VAX/VMS VNEWS 1.41

phsieh@lambert.waterloo.edu (Paul Hsieh) writes...
>>>>You are put in a room by evil Nazi's.
>                     If you take "Friendless"'s (aka putty brain) approach
>what's to say you don't have 99 other putzs doing the same thing?  The

what's to say you do?

>probability of living in such a scenario?  0.  Of course this seems to be a
>worse case scenario but on the other hand if you intend to live with such an
>approach, you are DEPENDING on the fact that at least someone is not going to
>follow this approach.  

I take it you have a strategy which DOESN'T depend on that fact!?

>So what to do?  We've seen from above that symmetry will cause this (and any
>other) fixed strategy to fail.  

There is no symmetry principle here.

>You might argue that the first "attempt" would fair pretty well (even better!)
>if the other 99 people were to use this probabilistic strategy.  But then
>again think of it from that persons perspective.  There is no way that s/he can
>ever know for sure what strategy other people are employing and hence cannot
>conclude that people will use a strategy that is different from him/her.

Neither can she conclude they are using the same strategy as her.

>is a hybrid of the two approaches?  I.e., chose another probability q, and with
>probability q you will employ the p probabilistic approach, otherwise you will
>simply not press the button.  But when I worked this out, it ended up just
>being the p probabilistic method in disguise.

This should be obvious.

>Of course I have not really given a proof of the fact that my approach is
>"best" and in fact I am not entirely sure that it is.

It is not. Mr Friendless' strategy is better. You have made some faulty
assumtions about 'intelligent' people and optimal strategies.

Dave Ring
dwr2560@zeus.tamu.edu
From: uslave@mcl.mcl.ucsb.edu (KJS @ucsb.mcl)
Newsgroups: rec.puzzles
Subject: Re: Monday morning brain energizer April 6th
Keywords: A weekly series in the making?
Date: 7 Apr 92 05:54:25 GMT
Distribution: na

In <1992Apr6.160440.28526@clipper.ingr.com> kontos@clipper.ingr.com (Thorne Kontos) writes:


>        The Deception Test

>A,B,C,D, and E have just finished taking a test for Deception.
>In their usual chatty way, they are making remarks about their
>places. The remarks of the first and second place finishers
>are false, while the others are true. Lets listen in now.

>A: "D was third"
>B: "E was not first"
>C: "I was not last"
>D: "C was lower than B"
>E: "B was second"

>What were their places?


* SPOILER ALERT!!! *

First off, E MUST be first, since he can't be telling the truth about being
second, or lying about being third-fifth.  That means B was telling the truth,
so he's third-fifth.  C cannot be second, since he'd be telling the truth, so
he's also third or fourth, since he wasn't last.  If A was telling the truth,
D's statement would be false.  That wouldn't be possible if he was third like
A said, so A must be lying.  That makes A second, B third, C fourth, and D
fifth.

So, in order, it's E, A, B, C, D.


--
*        *      *     *     *    *   *  *  * * ** ** *** **** ****** ********* 
   KJS @ucsb.mcl (Jesse Chang)               e-mail: uslave@mcl.mcl.ucsb.edu *
   "Traveling thru hyperspace ain't like dusting crops, boy!"    -- Han Solo * 
*         *       *      *     *    *   *  *  * ** ** *** **** ****** ********
From: monson@diablo.amd.com (Steve Monson)
Newsgroups: rec.puzzles
Subject: Re: farms
Date: 8 Apr 92 20:45:59 GMT
Organization: Advanced Micro Devices, Inc., Austin, TX

>    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?

Well, with the obvious associations, we have

	h + g + c = 100
and   10h + g + (c/8) = 100,
 or   80h + 8g + c    = 800

subtracting the top equation from the bottom one, we have

      79h + 7g = 700

Now, we can't have very many horses, and since 79 isn't divisible by 7,
let's assume 7 horses. That gives us

      79*7 + 7g = 553 + 7g = 800
so
	h=7 ==> g=21 ==> c=72

And,	7 + 21 + 72 = 100
       70 + 21 + 9  = 100

And that's that.

Steve
-- monson@diablo.amd.com
	The tide gone out for good
	Thirty-one words for seaweed		"The Death of Irish"
	Whiten on the foreshore.		by Arden Carl Mathew
Newsgroups: rec.puzzles
From: macster@IASTATE.EDU (Michael C Mccarty)
Subject: Re: A question of time
Reply-To: macster@IASTATE.EDU (Michael C Mccarty)
Organization: Iowa State University
Date: Wed, 15 Apr 1992 08:26:00 GMT

In article <mblum.703293937@chaos.cs.brandeis.edu>, mblum@chaos.cs.brandeis.edu
(Cerebus) writes:
> ncramer@bbn.com (Nichael Cramer) writes:

   [...stuff deleted...]

>                 
> 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
>

I'm with you... I don't like it much either.  It sort of begs the question,
"Well, was the clock actually correct?" (i.e. How can you say it did if it
didn't read the correct time?)  I was thinking, though I haven't come up with a
solution yet, that possibly, since he had blurry vision, that he accidentally
switched the hands of two analog clocks.  Once, he reads the correct time and
another he reads an almost mirror-image of the time.  At the moment, I can only
get it to advance 1 hr. and 5 min., but I'll see what I can do.

				Just Thinking,
				Mike
Newsgroups: rec.puzzles
From: earutter@athena.cas.vanderbilt.edu (Ethan Rutter)
Subject: Re: farms <SPOILER>
Keywords: gch
Reply-To: earutter@athena.cas.vanderbilt.edu (Ethan Rutter)
Organization: Mathematics, Vanderbilt University, Nashville
Date: Fri, 10 Apr 92 23:19:08 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?

Let h = # horses bought, g = # goats bought, and c = # chickens bought.
Assume c is divisible by 8 (chickens can only be bought in groups of 8).

Then 10 h + g + c/8 = 100 & h + g + c = 100.
Then c = 72/7 h.  So 7n horses are bought, else c mod 8 is not 0.
Now if n > 1, then c > 144 which can't be.  
So h = 7, g = 21, and c = 72.  


-- 
-------------------------------------------------------------------------
Ethan Rutter                       | "If I claim to be a wise man,
earutter@athena.cas.vanderbilt.edu |  it surely means that I don't know."
Math Dept.                         |      - Carry On Wayward Son, by Kansas
Newsgroups: rec.puzzles
Subject: Re: The Foreign Counterfeiter
From: chriscasey@otago.ac.nz
Date: 13 Apr 92 21:49:22 +1300
Organization: University of Otago, Dunedin, New Zealand

In article <YdtPlVK00iV1M2Gc8M@andrew.cmu.edu>, ss8c+@andrew.cmu.edu (Simon Szykman) writes:
> 
> Fred is a counterfeiter in a foreign country.  He manages
> to steal two printing plates for local paper currency, for
> his "printing" shop.  The local currency is called bux, and
> he the plates he has are for m bux notes and for n bux notes
> (where m and n are integers).  He can therefore have a limitless
> supply of m and n bux notes.
> 
> Fred also has a really bad guilt complex about printing up
> this funny money.  If he tries to buy something where he'd
> get change back, the idea of the person giving him real money
> back, when he's paying with fake money, causes him to confess.  
> 
> Fred is also really cheap, so he would NEVER say anything like
> "keep the change."
> 
> Therefore, when he buys things, he only buys things he can get
> for exact change.  The good thing is that everything in this
> country costs an integer amount of bux, including tax, so that 
> he can find things to buy for exact change often enough.
> 
> 1.    If and only if m and n have a certain property, there is
> a number X, such that Fred can buy ANYTHING that costs more than
> X bux.  What is that property? 
> 
> 2.    What is X?
> 
>  
>                             - Simon           
> 
As long as at least one of m and n is unity, then Fred can buy anything
regardless of amount - therefore X = 0 [or -1, if there are giveaways! :-) ]. 
This seems a little trivial - I think I've missed something...?

Maybe, as long as m and n have mixed parity, then X = Max{m,n} - 1.


Chris
From: ss8c+@andrew.cmu.edu (Simon Szykman)
Newsgroups: rec.puzzles
Subject: The Foreign Counterfeiter
Date: 10 Apr 92 16:03:45 GMT
Organization: Masters student, Mechanical Engineering, Carnegie Mellon, Pittsburgh, PA


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

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

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

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

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

2.    What is X?

 
                            - Simon           


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

What is the next number in this series?

1, 11, 21, 1211, 111221, ?
Newsgroups: rec.puzzles
From: surya@watson.ibm.com (Surya Mohanty)
Subject: Re: Primes in sequences...
Date: Mon, 13 Apr 1992 17:56:01 GMT
Reply-To: surya@rothko.watson.ibm.com (Surya Mohanty)
Disclaimer: This posting represents the poster's views, not necessarily those of IBM
Organization: IBM T.J. Watson Research Center

Q2 has already been asked . The anser is thre is exactly one prime in
the sequence 1, 101, 10101, .... ( that is 101).
1, 11, 111, ...
It is still an open problem whether the above sequence has finite or
infinite primes.
Newsgroups: rec.puzzles
From: wyg9633@uxa.cso.uiuc.edu (Wooed Sun)
Subject: Re: Math Problem
Organization: University of Illinois at Urbana
Date: Wed, 8 Apr 1992 05:57:16 GMT

amorgan@Xenon.Stanford.EDU (Crunchy Frog) writes:

>In article <CU311HL@math.fu-berlin.de> 
>  guckes@math.fu-berlin.de (Sven Guckes) writes:
>>BANDA@AUSVM1.VNET.IBM.COM ("Venu Prabhakar Banda") writes:
>>
>>>Problem: Show that (4444 ** 4444) is divisible by 9
>>>
>>>Cheers,
>>>
>>>Venu
>..[deleted].. I don't,
>however, think that 4444**4444 *is* divisible by 9 (evenly, that is).
>It can only be divisible by 9 if it is divisible by 3, which can only
>be if 4444 is divisible by 3.  It isn't.

>What am I missing?
>C Frog

I think you are very right.
Could we ask the original poster how (s)he got a misconception that
the number would be divisible by 9.
Of course, I assume the '4444' was meant to be a decimal number.

-- 
email : w-yang@uiuc.edu (wyg9633@uxa.cso.uiuc.edu)
From: morris@anasaz (Bob Morris)
Newsgroups: rec.puzzles
Subject: Re: Isaac Asimovs mysteries puzzle (hint and comment)
Date: 9 Apr 92 17:27:11 GMT
Organization: Anasazi, Inc.  Phoenix, Az

In article <3847@shum.huji.ac.il> yehuda@shuldig.huji.ac.il (Jonathan Berlinger) writes:
>Donn't know if anyone heard this one yet, so:
>
>Why does>	Christmas = Halloween
>
Comment:  This works in the US and some other places, but not everywhere.
And no, I don't mean places where they don't celebrate Christmas.  It
doesn't work in some places where Christmas is celebrated.  (Does anybody
outside of the US celebrate Halloween?)

Puzzle #1:  Why doesn't it work everywhere?  Where does it work?

Puzzle #2: Why does Christmas = Thanksgiving (sometimes but not always)

Puzzle #3: Why does Halloween = Thanksgiving (sometimes but not always)

(The answer to #2 and #3 is the same as the original.)

Since the original article was posted worldwide, I've followed up
worldwide, but further follups should probably be restricted to USA only.

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: csmith@onetouch.COM (Craig Smith)
Subject: Re: A simple sequence of numbers <SPOILER>...
Organization: MCS/OneTouch, Inc.
Date: Wed, 08 Apr 92 23:47:37 GMT

hemmady@Cadence.COM (Shankar Hemmady) writes:



>Continue this sequence:

>	3, 3, 5, 4, 4, 3, 5, 5, 4, 3, ...


one     3 letters
two     3 letters
three   5 letters
four    4 letters
five    4 letters
six     3 letters
seven   5 letters
eight   5 letters
nine    4 letters
ten     3 letters
eleven  6 letters
twelve  6 letters

etc.
-- 
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: Craig Smith                                         csmith@onetouch.COM ::
:: Columbus, OH                                  ...!uunet!onetouch!csmith ::
::::::::::::::: "Life is a fulltime occupation."--Gene Kalman :::::::::::::::
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 12:53:31 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: lordSnooty@cup.portal.com (Andrew - Palfreyman)
Newsgroups: rec.puzzles
Subject: Re: Isaac Asimovs mysteries puzzle
Date: 7 Apr 92 03:25:48 GMT
  <1992Apr6.184219.18452@murdoch.acc.Virginia.EDU>
Organization: The Portal System (TM)

isaac asimov died today of kidney failure at the age of 72.
From: jmhodapp@aplcomm.jhuapl.edu (Jon Hodapp)
Subject: Re: Afterwards
Organization: JHU/APL
Date: Tue, 14 Apr 92 16:22:56 GMT

In article <3283@inca.comlab.ox.ac.uk>, sharkey@robots.ox.ac.uk (Paul Sharkey) writes:
> 
> Afterwards is a 10 letter word that can be "touch-typped" using one
> hand. Can anyone better that?
> 

John Hopkin U
(all right, I don't know if shift and space count, they are on the right of my
keyboard, and the necessary 's''s are on the left)

Jon               JMHODAPP@aplcomm.jhuapl.edu
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When all else fails, try X=8     - Tom Hanks, on Family Ties




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