HDTenEightyP/NetNews huggingface dataset:
From: hook@wk01.nas.nasa.gov (Edward C. Hook)
Subject: Re: order of numbers?
Organization: NAS, NASA Ames Research Center, Moffett Field, CA
Date: Tue, 7 Apr 92 13:47:20 GMT
Summary: Is this a spelling flame ?
Distribution: rec

In article <9385@wrgate.wr.tek.com> dant@logos.wr.tek.com (Dan Tilque) writes:
>pfalstad@phoenix.Princeton.EDU (Paul Falstad) writes:
>>dant@logos.wr.tek.com (Dan Tilque) wrote on 6 Apr 92 17:01:18 GMT:
>
>>>	8, 3, 1, 5, 9, 7, 6, 4, 2
>>I'll guess German, based on where the "2" is.
>
>Right again.

-- This doesn't SEEM right - if it's German, then '7' == 'sieben', '6' == 'sechs'
   so the order would be

            8,3,1,5,9,6,7,4,2

   ...

>
>---
>Dan Tilque	--	dant@logos.WR.TEK.COM

===============================================================================
 Ed Hook                                  |  "I have yet to see any problem,
 Computer Sciences Corporation            |   however complicated, which, when
 NAS                                      |   you looked at it in the right 
 NASA/Ames Research Center                |   way, did not become still more
 Mountain View, CA 94039                  |   complicated." - Poul Anderson
 +1 415 604 1272                          |-----------------------------------
 Internet: hook@nas.nasa.gov              | They're MY opinions ... ALL mine!!
===============================================================================
From: bas@phys.uva.nl (Bas de Bakker)
Newsgroups: rec.puzzles
Subject: Re: order of numbers?
Date: 7 Apr 92 16:59:36 GMT
Distribution: rec
Organization: Institute for Theoretical Physics, Amsterdam, the Netherlands
In-reply-to: pfalstad@phoenix.Princeton.EDU's message of 6 Apr 92 19:05:47 GMTLines: 13

In article <1992Apr6.190547.27198@Princeton.EDU> pfalstad@phoenix.Princeton.EDU (Paul Falstad) writes:


>   >	8, 3, 1, 5, 9, 7, 6, 4, 2

>   I'll guess German, based on where the "2" is.

I don't know what the language is, but it's not German. In German it
would be:

8, 3, 1, 5, 9, 6, 7, 4, 2  (almost the same though)

Bas de Bakker.
Newsgroups: rec.puzzles
From: thssrkr@iitmax.iit.edu (Raghunath K Rao)
Subject: Re: Circular number puzzle
Organization: Illinois Institute of Technology
Date: Mon, 6 Apr 92 21:22:54 GMT

In article <702434144.F00001@csource.oz.au> Mick.Gulovsen@f503.n635.z3.csource.oz.au (Mick Gulovsen) writes:
>Hi there Shere,
>
>At last an easy one
>
>
>
> S>               ABCDEF * 2 = CDEFAB
> S>               ABCDEF * 3 = BCDEFA
> S>               ABCDEF * 4 = EFABCD
> S>               ABCDEF * 5 = FABCDE
> S>               ABCDEF * 6 = DEFABC
>
>By reasoning A MUST be 1 to keep the result of the x 6 calc to a 6 digit No.
>After that it is a simple matter to get the rest (ie x3 ends with a 1 MUST mean
>F=7 etc).
>Here is the result
>
>A=1
>B=4
>C=2
>D=8
>E=5
>F=7
>
>CYA
>
> * Origin: Andy's BBS: Melb Australia - 61-3-359-6378 (3:635/503)

Generates the 6 digit repeating sequence of 1/7 = 0.142857142857.

---------------------------------------------------------------------
                                  _____         ___        _______
K. RAGHUNATH RAO                 /____/ \      /__/\      /______/\
email : thssrkr@iitmax.iit.edu  |  __ \ /|    / _ \ \    /  ___  \ |
sleepy   (res) : (312)791-9428  | |__) |/    / /_\ \ \   | / | \ | |
sleepier (off) : (312)567-3407  | __  /\    |  ___  | |  | | | | | |
                                | | \ \ \   | / | \ | |  | | |_| | |
It is the intonation and not    | | |\ \ \  | | | | | |  | \/__/ |/
the intention that matters!!    |_|/  \_\/  |_|/  |_|/    \______/ 

---------------------------------------------------------------------
Xref: sparky rec.puzzles:3540 soc.culture.indian:13674
From: surya@watson.ibm.com (Surya Mohanty)
Newsgroups: rec.puzzles,soc.culture.indian
Subject: Re: A (Not So Obviously) simple number puzzle
Date: 8 Apr 92 18:40:18 GMT
Reply-To: surya@rothko.watson.ibm.com (Surya Mohanty)
Organization: IBM T.J. Watson Research Center
Disclaimer: This posting represents the poster's views, not necessarily those of IBM

The answer to the question certainly depends on the unit of time. 
If the unit of time is year then (27, 24, 3) is a valid answer. 
Change the unit to months (256, 252, 4) would then be a valid answer. 
Different solutions can be found for unit being hours, days, minutes,
seconds and micro-seconds. (ILL POSED QUESTION ).
Newsgroups: rec.puzzles,soc.culture.indian
From: russotto@eng.umd.edu (Matthew T. Russotto)
Subject: Re: A simple number puzzle
Date: Wed, 08 Apr 92 02:16:47 GMT
Organization: College of Engineering, University of Maryland, College Park

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?"

child	wife	you
1	0	1 	naa
2	2	4	probably not
3	24	27	sure
4	252	256	naa

Sometimes, there is no substitute for exhaustive search.


-- 
Matthew T. Russotto	russotto@eng.umd.edu	russotto@wam.umd.edu
Some news readers expect "Disclaimer:" here.
Just say NO to police searches and seizures.  Make them use force.
(not responsible for bodily harm resulting from following above advice)
Organization: Freshman, Math/Computer Science, Carnegie Mellon, Pittsburgh, PA
Newsgroups: rec.puzzles
Date: Tue, 14 Apr 1992 18:08:32 -0400 
From: "Daniel S. Efran" <de0t+@andrew.cmu.edu>
Subject: Re: Game Theory

> [...] 4) asking a dummy
>question (ie asking for a card which is already face up 
>on the table) [...]

Why would you do that?  Isn't it just a waste of a turn?
In fact, I'm surprised that the game allows it at all --
it seems equivalent to skipping a turn, or 'passing.'  If
that's allowed, why not just say 'Pass'?  If not, 'dummy
questions' allow you to cheat.

Am I missing something?

 
	---Daniel Efran
----------------------------------
An it harm none, do as you Will.
Practice random kindness and senseless acts of beauty.
From: baustad@idt.unit.no (Jostein Baustad)
Newsgroups: rec.puzzles
Subject: Re: C-puzzle (Tough)
Keywords: puzzle
Date: 8 Apr 92 09:05:54 GMT
Reply-To: baustad@idt.unit.no (Jostein Baustad)
Organization: Norwegian Institue of Technology

In article <1992Apr7.125412.18567@fwi.uva.nl>, mdeboer@fwi.uva.nl (Marc de Boer (I88)) writes:
|> 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)
|> --------------------------------------------------------------------

My suggestion is that the first program is a newsarticle-duplicator (at least
we received three copies of your article!) and that the second one is a
.signature-duplicator. Correct??

Jostein.

-- 
*******************************************************************************
* Jostein Baustad                    * E-mail : baustad@idt.unit.no           *
* Norwegian Institute of Technology  *          baustad@solan.unit.no         *
* Division of CS and Telematics      * "Reality is just a convenient          *
* Information and Knowledge Systems  *  measure for complexity."              *
*******************************************************************************
*             Queen's Park Rangers F.C. - SuperHoops !                        *
*******************************************************************************
From: steiner@bgsu.edu (Ray Steiner)
Newsgroups: rec.puzzles
Subject: Re: Tough Cryptics
Summary: Description of magazine
Date: 17 Apr 92 14:18:55 GMT
Organization: Bowling Green State University B.G., Oh.


The first issue of TOUGH CRYPTICS had 4 plain American cryptics
(a la Dell mags) and a good variety puzzle by Cox and Rathvon.
Future issues will have 4 plain and 3 variety cryptics. Each
issue has 2 clueing contests. I liked the first issue a lot
and am looking forward to more of the same!
Ray Steiner
From: uslave@mcl.mcl.ucsb.edu (KJS @ucsb.mcl)
Newsgroups: rec.puzzles
Subject: Re: Dying young
Date: 7 Apr 92 06:30:40 GMT

In <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?

If I push the button, I die.  If I don't, I could either live or die, depending
on what the other 100 people do.  If on one else pushes his button, I die
anyway, so why remove my chance of living by killing myself?  I don't press
the button, and hope that someone else does.  Of course, if everyone else
thought this way, then we're all dead.  Plus, we're captured by EVIL NAZIS
who were probably lying and are planning to kill us no matter what, so I'm
dead no matter what.  :-)



--
*        *      *     *     *    *   *  *  * * ** ** *** **** ****** ********* 
   KJS @ucsb.mcl (Jesse Chang)               e-mail: uslave@mcl.mcl.ucsb.edu *
   "Traveling thru hyperspace ain't like dusting crops, boy!"    -- Han Solo * 
*         *       *      *     *    *   *  *  * ** ** *** **** ****** ********
From: jerryd@hpgrla.gr.hp.com (Jerry Donovan)
Date: Tue, 14 Apr 1992 14:48:26 GMT
Subject: Re: Brick Problem (crude ASCII)
Organization: Hewlett-Packard, Greeley, CO
Newsgroups: rec.puzzles

>>>Hence, counting the outside space as another "brick", such puzzles can
>>>only be solved without a trick if the number of bricks with odd numbers
>>>of walls is 2 or 0. (In the latter case both ends of the line must be in
>>>one brick.) 

>jerryd@hpgrla.gr.hp.com (Jerry Donovan) writes...
>>1 will also work.

>Hmmm... the total number of brick-sides is odd... now that's a good trick,
>considering that each wall has two sides. ;)

Yea, I skimmed the previous post too quickly, and missed the phrase 'counting
the outside space as another "brick"'.  Previous post was 100% correct.

>Dave Ring
>dwr2560@zeus.tamu.edu

Jerry (need to slow down and actually read these posts) Donovan
From: hxie@dps91a.cis.upenn.edu (Hong Liang Xie)
Newsgroups: rec.puzzles
Subject: Re: Ladder puzzle (??)
Date: 17 Apr 92 18:37:24 GMT
Organization: University of Pennsylvania

In article <1992Apr17.175421.12409@lambda.msfc.nasa.gov> holt@lambda.msfc.nasa.gov writes:
>
> I have an interesting little puzzle.
>
>     -   |       |
>     |   |\     /|  -
>     |   | \   / |  |
>         |  \ /  |
>    15   |   x   |  10
>         |  / \  |
>     |   | / 6 \ |  |
>     |   |/  |  \|  |
>     -   ---------  -
          | x | y |

Assume the distance from the intersection to left wall 
is x, to the right wall is y.  

(x+y) could be any value so long as x/y = 3/2.  

Do I miss anything?  

  




Newsgroups: rec.puzzles
From: hemmady@Cadence.COM (Shankar Hemmady)
Subject: A simple sequence of numbers ...
Organization: Cadence Design Systems, Inc.
Date: Tue, 7 Apr 1992 22:07:10 GMT



Continue this sequence:

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


-Shankar Hemmady
hemmady@cadence.com
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Newsgroups: rec.puzzles
Subject: Re: Math Problem
Date: 11 Apr 92 22:28:00 GMT
Organization: Texas A&M University, Academic Computing Services
News-Software: VAX/VMS VNEWS 1.41

clong@romulus.rutgers.edu (Chris Long) writes...
>>> sod ( sod ( sod (4444 ^ 4444 ) ) )
>We've all seen this one before ... now, what is the largest x for
>which we can easily compute sod^3(x)?

Well, the question is kind of silly when put that way. But if 'easily'
means using only the easiest way to solve the original problem, I think
the answer is x=9*10^[10^9-2]-1. Then s(x)<8999999992, ss(x)<82, sss(x)<16
sss(x)=8.

Dave Ring
dwr2560@zeus.tamu.edu
Newsgroups: rec.puzzles
From: karabin@triton.cis.ohio-state.edu (Eric Karabin)
Subject: Re: The Homing Pigeon Problem and more...
Originator: karabin@triton.cis.ohio-state.edu
Organization: Ohio State Computer Science
Date: Mon, 6 Apr 1992 18:56:29 GMT

In article <1992Apr6.023817@IASTATE.EDU> macster@IASTATE.EDU (Michael C Mccarty) writes:

   Newsgroups: rec.puzzles
Path: cis.ohio-state.edu!zaphod.mps.ohio-state.edu!rpi!uwm.edu!ux1.cso.uiuc.edu!news.iastate.edu!IASTATE.EDU!macster
   From: macster@IASTATE.EDU (Michael C Mccarty)
   Sender: news@news.iastate.edu (USENET News System)
   Reply-To: macster@IASTATE.EDU (Michael C Mccarty)
   Organization: Iowa State University
   Date: Mon, 6 Apr 1992 07:38:17 GMT
   Lines: 42


	   -Suppose there are two armies of soldiers engaged in a war against a common
   enemy.  These armies are separated by mountain range, thereby making homing
   pigeon the only form of communication between them.  The commander of Army A
   knows the position of the enemy, of which, both armies are able to get to.
   However, the enemy is so strong that if both armies do not attack
   simultaneously, it would mean certain disaster for both armies.  The commander
   of Army A wishes to send Army B a message stating a time at which both armies
   may attack.  Unfortunately, the homing pigeons are a rather unreliable form of
   communication (they tend to get lost, killed by predators, etc.).  Therefore
   Commander A requests that Commander B send a reply message stating the approval
   of the plan.  Commander B also realizes the limitations of the homing pigeon,
   and requests a return message acknowledging his message.  Well I think you can
   see where this is leading...  What message, if any, can the commanders send to
   each other that will ensure that an answer to whether the message was received
   or not can be implied?

   I, honestly, do not know a definite answer to this problem.  I thought maybe if
   you people worked on it, that we all may gain a little insight into the
   solution.



	I just recently completed a course in Computer Networking here at Ohio
State.


	The above paragraph is not a puzzle...  It is a mere example of why
you should use a Three-Way Handshaking algorithm when designing a protocal.

	A sends to B,
	B sends to A,
then,   A sends to B.

		This is obviously capable of being a never-ending situation.

BUT, it has been proven that Three-Way Handshaking is more than satisfactory
for most any communication.   (Yes, problems can still occur)


	So, in the worst case scenario:

		YOUR ARMIES ARE DEAD!!!!!!!!
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Newsgroups: rec.puzzles
Subject: Re: Fill in the blanks.
Date: 11 Apr 92 20:07:00 GMT
Followup-To: rec.puzzles
Organization: Texas A&M University, Academic Computing Services
News-Software: VAX/VMS VNEWS 1.41

In article <1992Apr10.203558.34277@watson.ibm.com>, surya@rothko.watson.ibm.com (Surya Mohanty) writes...
>What are the next three numbers in the sequence and why ?
> 
>3 5 11 17 31 41 67 - - - ?

Did you forget 59?

Dave Ring
dwr2560@zeus.tamu.edu
Newsgroups: rec.puzzles
From: hemmady@Cadence.COM (Shankar Hemmady)
Subject: Re: Circular number puzzle (SPOILER)
Organization: Cadence Design Systems, Inc.
Date: Tue, 7 Apr 1992 22:48:47 GMT

In article <702434144.F00001@csource.oz.au> Mick.Gulovsen@f503.n635.z3.csource.oz.au (Mick Gulovsen) writes:
>Hi there Shere,
>
>At last an easy one
>
>
>
> S>               ABCDEF * 2 = CDEFAB
> S>               ABCDEF * 3 = BCDEFA
> S>               ABCDEF * 4 = EFABCD
> S>               ABCDEF * 5 = FABCDE
> S>               ABCDEF * 6 = DEFABC
>
>By reasoning A MUST be 1 to keep the result of the x 6 calc to a 6 digit No.
>After that it is a simple matter to get the rest (ie x3 ends with a 1 MUST mean
>F=7 etc).
>Here is the result
>
>A=1
>B=4
>C=2
>D=8
>E=5
>F=7
>
>CYA
>
> * Origin: Andy's BBS: Melb Australia - 61-3-359-6378 (3:635/503)


It is also interesting to observe that the number 142857 is the recurring
sequence that you obtain for 1/7

1/7 = 0.142857 142857 ...

Now 2/7, 3/7, 4/7, 5/7, 6/7 correspond to the recurring circular sequences which
are the same numbers in the problem.


-Shankar Hemmady
From: miller@lclark.edu (John Miller)
Newsgroups: rec.puzzles
Subject: Re: Security Locks
Date: 14 Apr 92 17:24:51 GMT
Article-I.D.: lclark.1992Apr14.172451.25946
Organization: Lewis & Clark College, Portland OR

rgerbas@csws16.ic.sunysb.edu (Richard S Gerbasi) writes:
>OK everyone, I hope this one isn't two much for this newsgroup.  

What a jerk!

>In one of Applied Math classes, we were assigned this problem as an
>extra credit problem.  Due to a growing number of tests I have to take,
>I figured I would post it here for some advanced insight on the problem
>and perhaps a solution. By solution I mean fairly detailed so I can 
>follow it.  Oh, and by the way if my Professor is reading, I promise not
>to use any form of solution given to me from this Newsgroup.  (Well
>maybe not promise, but I will try!!!)  Well here it is!

This or a similar problem was enumerated recently in alt.security.  Perhaps
you could check the archive there.

Personally, I might have responded having gotten past your opening insult,
and had you spared me the explanation of "advanced insight".  Others who
have asked for help preparing for a programming contest and the like have
been flamed for being blatantly unethical.  What will you do next due to
a growing number of tests to take?

-- 
miller@lclark.edu					"What's ahead?"
Newsgroups: rec.puzzles
From: guckes@math.fu-berlin.de (Sven Guckes)
Subject: Re: A simple sequence of numbers ...
Originator: guckes@medusa
Organization: Free University of Berlin, Germany
Date: Wed, 8 Apr 1992 16:06:59 GMT

hemmady@Cadence.COM (Shankar Hemmady) writes:
>Continue this sequence:
>	3, 3, 5, 4, 4, 3, 5, 5, 4, 3, ...
	6, 6, 8, 8, 7, 7, 9, 8, 8, 6, ...

BEEP! My FAQ-o-meter just went off. Somebody please add this puzzle
and the reference to that Book Of Sequences to the FAQL. Thank you!
Newsgroups: rec.puzzles
From: amorgan@Xenon.Stanford.EDU (Crunchy Frog)
Subject: Re: primes in 1, 11, 111, etc.
Organization: Computer Science Department, Stanford University.
Date: Thu, 16 Apr 1992 07:25:44 GMT

In article <PALMERP.92Apr15220643@ilanga.math.orst.edu> 
  palmerp@math.orst.edu writes:
>In article <sa4h9INNrt5@agate.berkeley.edu> 
>  davidb@triangle.Berkeley.EDU (David T. Blackston) writes:
>
>   The question of whether the number of primes consisting of all ones
>   in their decimal representation is finite is unsolved.  
>
>   Here are some interesting (?) related questions.  Let a_n be the 
>   number consisting of n ones.
>
>   Question 1: Show that a_n is prime only if n is prime.  (Easy)
>
>I came into this discussion late, so maybe I'm confused. I wrote a 1-line
>Mathematica program to factor a_n through a_41, and the only primes it found
>were a_19 and a_23. Here's the program in case someone with Mathematica wanst
>to check it out. 

.. deleted ..

>Am I just tired/confused?

Confused.  David said a_n is prime *only* if n is prime.  Another way
of phrasing this is "if a_n is prime, then n is prime"

>Paul Palmer
>Department of Mathematics     	E-mail: palmerp@math.orst.edu

C Frog
From: dant@logos.wr.tek.com (Dan Tilque)
Newsgroups: rec.puzzles
Subject: Re: order of numbers?
Date: 6 Apr 92 17:01:18 GMT
Distribution: rec
Organization: Upper Left Coast Liberation Army

dik@cwi.nl (Dik T. Winter) writes:
>dant@logos.wr.tek.com (Dan Tilque) writes:
> > dik@cwi.nl (Dik T. Winter) writes:
> > > > dm@stekt2.oulu.fi (Hannu Helminen ti) writes:
> > > > >8, 5, 4, 9, 1, 7, 6, 3, 2.
> > > > >But why are they in that particular order?
> > >But added points: why did he not give:
> > >	8, 2, 3, 6, 4, 7, 5, 9, 1
> > Perhaps for the same reason he didn't give:
> > 	2, 9, 8, 4, 5, 7, 6, 3, 1
>I do not think so.  Considering the original comes from Finland I would have
>thought that 823647591 is more natural than 854917632, assuming the poster
>is of Finnish origin of course.  (Hint: both list the numbers in alphabetic
>order.)

When I came up with the last of the above three lists, I was not
choosing numbers at random (although I made an error and put a seven in
the last position where there should have been a 1--corrected in this
article).  Question: what language is it?  (And what is zero in that
language?  Something I never learned.)

Here's another one:

	8, 3, 1, 5, 9, 7, 6, 4, 2

All four of these lists, 7 comes in the sixth position.  I wonder if this
is only a coincidence.

---
Dan Tilque	--	dant@logos.WR.TEK.COM
Newsgroups: rec.puzzles
From: surya@watson.ibm.com (Surya Mohanty)
Subject: Simple NT problem
Date: Mon, 13 Apr 1992 16:07:26 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

If a and b are positivite integers and if 
     1+ a^2 + b^2
k= --------------
       a*b
is also an integer then prove that k is equal to 3.
Newsgroups: rec.puzzles
From: sjs@hood.hood.caltech.edu (Stephen Speicher)
Subject: 2 oldies
Organization: California Institute of Technology, Pasadena
Date: Tue, 7 Apr 1992 15:55:27 GMT

These are two simple old ones - but cute.

1. Consider the sequence O T T F F . .
   What are the next two letters - and why?

2. Consider the letters H I J K L M N O
   There is a one-word solution to this.

Stephen
sjs@hood.hood.caltech.edu
From: killer@psykoman.cs.hut.fi (Kalle Kivimaa)
Newsgroups: rec.puzzles
Subject: Re: Fill in the blanks.(SPOILER?)
Date: 12 Apr 92 15:16:24 GMT
Organization: Helsinki University of Technology, CS-lab

In <11APR199215071585@zeus.tamu.edu> dwr2560@zeus.tamu.edu (RING, DAVID WAYNE) writes:
>In article <1992Apr10.203558.34277@watson.ibm.com>, surya@rothko.watson.ibm.com (Surya Mohanty) writes...
>>What are the next three numbers in the sequence and why ?
>>3 5 11 17 31 41 67 - - - ?
>Did you forget 59?

If he did, then the next numbers are:

83,109 and 127.  If you take all the primes and then start taking every
p:th prime where p itself is a prime, then you will get 2nd (3), 3rd (5),
5th (11) etc primes.
--
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