Daily puzzles/problems.

[MarkBastable]

inhabited

[prendrelemick]

Yes.

So, back to your question.







Which is the odd one out?

Milton, Heaney, Tennyson, Penn Warren, Betjeman, Frost

[Scheherazade]

Milton?

[MystyrMystyry]

Milton wasn't poet laureate

[MarkBastable]

Yep...

[billl]

Pedf dxa xwahv pehp d ehsa df jtdpa iqqk, otp dp'f pdva pq iap ohxr pq pea nhwhka qc ydca. Fqvapdvaf dp fpdmrf, otp gqt masaw rmqz zehp zdyk fptcc zdyy ehnnam hwqtmk pea mabp xqwmaw. Otp d yqsa pedf xwhn... D fphwa kqzm fptcc pehp xqtyk rdyy gqt dc dp cqtmk gqt. Zeawa hv d?

[MarkBastable]

I think the reference is beyond me. I have no idea what's implied by the ice cream.

[billl]

A clue will eventually come along, if necessary.

[billl]

What with this thread losing a bit of steam recently, my latest puzzle pretty much requiring a computer printout, and regulars being scattered across time zones, etc., it's hard to pace things well, as far as clues go.

Really, the best clue I could possibly give isn't in the puzzle at all.

[Scheherazade]

Is this based on a code, Bill? Ie, d=a, f=n etc?

[billl]

Yes, Scher, it's something like that, eventually revealing a mystery that managed to stop MarkBastable in his tracks.

[MarkBastable]

Is this based on a code, Bill? Ie, d=a, f=n etc?

I've done that bit. I just don't know what it means when it's translated.

Take the uppercase version to Word and replace uppercase 'in' with lowercase 'out' letters, starting with those in the last sentence which is a dead giveaway. Soon you'll have a para in English - which seems to allude to an ice cream situation I don't know about. But maybe I'm missing something.

[kasie]

I've got the paragraph de-coded but, like MB, don't know what it means. Is it a quotation from a book? If so, I might have an ide, but it's so long since I read it (the book I think it might be) that I'm not really sure.

[billl]

I hope the clue I gave helps! (Not from a book, it's an original riddle.)

[kasie]

A ringmaster in a circus during the interval? Nah, can't be - can it?

[billl]

VERRRRRYYYYY CLOSE, Kasie. (Did the clue help? Maybe not...?) Anyhow, pretty close....

But no.

[prendrelemick]

Are you in a zoo? or an animal park of some kind? Watching a circus parade? (do they still have them)

[kasie]

A policeman on traffic control taking a break? (Thinking how you suggested this might stop MB in his tracks....)

[billl]

I want to give the win to kasie, but mick has stumbled upon it. The only letter not in the puzzle's solution (and its encoded analogue is, of course, also missing in the encryption puzzle) is the letter "z".

Stopping MB in his tracks wasn't a clue, sorry about that. And the "parade of life" bit was a little more misleading than I should've made it, probably. "...on my parade through life," or something would've been better, but I was trying to eliminate unnecessary words because I didn't want it too long, and so that bit got a little weak.

Anyhow, I had imagined myself at the zoo, relaxing for a few minutes with some ice cream. Mick got it.

[prendrelemick]

Wh'd'ya mean stumbled!


what have the following in common.


Salmon
Island
Sword


In the unlikely event that no one gets it, I will add another each day.

[billl]

OK, you're right, you exhibited some skill when you got it in your sights, fair and square--but you took down two innocent scenarios with your scattershot blast.

[iamnobody]

Wh'd'ya mean stumbled!


what have the following in common.


Salmon
Island
Sword


In the unlikely event that no one gets it, I will add another each day.

Maybe not the answer you're looking for but, they all have silent letters.

[prendrelemick]

Correct.

Your go.

[iamnobody]

The following words have been placed in this order for a reason. What is it?

Trick
Squad
Serpent
Cathexis
Antiseptic
Octopus

[MarkBastable]

The following words have been placed in this order for a reason. What is it?

Trick
Squad
Serpent
Cathexis
Antiseptic
Octopus

The relative ages of the languages from which they originate?

(Total speculation. I'm much too lazy to do research.)

[iamnobody]

Not it.

[billl]

To stump us.

[iamnobody]

It would help to be thinking mathematically.

[jajdude]

The following words have been placed in this order for a reason. What is it?

Trick
Squad
Serpent
Cathexis
Antiseptic
Octopus

Tri, quad, pent, "six", sept, octo.

[iamnobody]

Sooo close, but you're off by one. "Six" is "hex". Close enough.

Your turn.

[prendrelemick]

^That's brilliant. I was barking up the completely wrong river.

[jajdude]

Ok, how about detecting something notable in this sentence:

One time the forest felt strange.

[MarkBastable]

Ok, how about detecting something notable in this sentence:

One time the forest felt strange.

It's an anagram of another sentence that means exactly the same thing.

[prendrelemick]

we're back to one, two, three, four, five, six.

[jajdude]

Yeah, I was lazy. If anyone can come up with that anagram I'd be impressed though.

[Scheherazade]

Truth be told, I have lots interest in the thread because some questions do not seem to offer enough mental excitement for me; I like it when I reach for pen and paper, so to speak, and work something out.

[prendrelemick]

Strange felt the forest one time.

There's your anagram:biggrin5:



Next one.

Everything was hunky dory until the phone rang -not one of my pet sounds at that time in the morning- like a dummy, I grabbed it off the wall. Hi! What's going on? I said in my best Steve McQueen voice. The Woman on the other end sounded different class. “You gotta help me” she said. I held the phone closer, that voice had more raw power than a loaded revolver. “My place has been ransacked, some of my vinyl is missing, but I don't know which ones.” It's a sign o' the times doll,” I drawled, “but nevermind, I'm on the case!”


There were at least 13 albums missing from her record collection. All top one hundred greatest of all time stuff, how many can you name? Bonus points for the artists.


Sorry scher no pencil needed.

[MystyrMystyry]

First three:

Hunky Dory - David Bowie

Pet Sounds - The Beach Boys

Revolver - The Beatles


Is it singles as well? Because Prince's Sign O' The Times, Off the Wall by Michael Jackson, What's Going On? by Four Non-Blondes, and few that seem to be obscured

[MANICHAEAN]

What's the speed of dark?

[MystyrMystyry]

Slightly faster than the speed of light - whenever light gets there, dark has already arrived and sitting comfortably on the couch with a stein of beer

[MarkBastable]

Dummy - Portishead

Steve McQueen - Prefab Sprout

Different Class - Pulp

Closer - Joy Division

Loaded - Velvet Underground

Raw Power - Iggy and the Stooges

Nevermind - Nirvana


Plus MM's three (Hunky Dory, Pet Sounds and Revolver) - and Sign o' the Times and Off the Wall are in fact both albums, as is What's Going On by Marvin Gaye.

[billl]

I couldn't spot anymore. (The was a Bangles album called "Everything", but, uh...)

[MarkBastable]

These can be grouped in threes - which is the one left over?

rain breast eye blood
mist haze peel
fever hammer
tit belly patch balls

[billl]

These can be grouped in threes - which is the one left over?

rain breast eye blood
mist haze peel
fever hammer
tit belly patch

I don't know if I am misunderstanding the puzzle, or if there was meant to be an additional word (or two less words) in the list. The way I saw it, since there are twelve words in the list, I was able to make four groups of three, with no words left over. Therefore, I have assembled three groups of three, as well as a fourth group of words that would not fit in the first three groups.

rain tit mist (words containing the letter "i" once)
eye peel fever (words containing the letter "e" twice)
breast haze hammer (words containing the letters "e" and "a" once each)
blood belly patch (the left overs)

Again, my groupings are probably all wrong ("rain" having an "a" in it, unlike its relatives, is particularly suspicious), I'm just throwing it out there.

[MarkBastable]

Crap - I missed one out. See edited original post.

Yeah, you need four groups of three which will leave one over.

Apologies, bill.



I don't know if I am misunderstanding the puzzle, or if there was meant to be an additional word (or two less words) in the list. The way I saw it, since there are twelve words in the list, I was able to make four groups of three, with no words left over. Therefore, I have assembled three groups of three, as well as a fourth group of words that would not fit in the first three groups.

rain tit mist (words containing the letter "i" once)
eye peel fever (words containing the letter "e" twice)
breast haze hammer (words containing the letters "e" and "a" once each)
blood belly patch (the left overs)

Again, my groupings are probably all wrong ("rain" having an "a" in it, unlike its relatives, is particularly suspicious), I'm just throwing it out there.

[billl]

No problem! Well, actually, I think I'll probably have a big problem figuring it out.

[Scheherazade]

Fever hammer?

[MarkBastable]

Fever hammer?

One left over - so, fever or hammer. Which is it? And why?

Don't be misled by the layout either. It could as easily have been this....

rain breast eye blood mist haze peel fever hammer tit belly patch balls

And the order they're given here doesn't matter either.

[billl]

All of the words will remain English words after having one or more letters removed from them, except for the word "eye".

Groupings:
fever ever, tit it, peel eel (remove first letter)
mist is, blood loo, balls all (remove first and last letters)
breast beast, rain ran, patch path (remove one interior letter)
haze he, hammer her, belly by (remove more than one interior letter)

NOTE: "breast" could also, unfortunately, fit into the last grouping (thus yielding "best"), and so this final grouping finally makes my suggested system here look a little less elegant than the first three groupings seemed to be promising.

[MANICHAEAN]

What happens if you get scared half to death, twice?

[Cunninglinguist]

What happens if you get scared half to death, twice?

You void you bowels.

[MarkBastable]

All of the words will remain English words after having one or more letters removed from them, except for the word "eye".

Groupings:
fever ever, tit it, peel eel (remove first letter)
mist is, blood loo, balls all (remove first and last letters)
breast beast, rain ran, patch path (remove one interior letter)
haze he, hammer her, belly by (remove more than one interior letter)

NOTE: "breast" could also, unfortunately, fit into the last grouping (thus yielding "best"), and so this final grouping finally makes my suggested system here look a little less elegant than the first three groupings seemed to be promising.


So much thought appears to have gone into that it seems almost churlish to say, No, not even close.

[prendrelemick]

I have a strong feeling its hammer. Now to work out why.



I'm thinking colours

[MarkBastable]

It's not hammer.

I may have overcomplicated it. I can make it simpler by cutting it down to these, primarily...

breast eye blood mist peel fever hammer tit belly balls

[prendrelemick]

Its peel, which is orange.

haze, patch, rain. purple
blood, tit, balls?. blue
fever, belly, hammer yellow
mist, breast, eye. red

[MarkBastable]

Yep. Well done.

An American will happen by any minute to explain 'blue balls'....

[prendrelemick]

I'm not sure I want to know.


Here's one for Scher. I copied it, because I wouldn't know where to begin to do one of these. I don't know if its hard or easy.




http://i85.photobucket.com/albums/k78/prendrelemick/mathsquiz.jpg

[billl]

Does anyone here have experience with these? Maybe there's a name for them...

EDIT: removed a rambling and roundabout exploration of the question of "operator precedence (http://en.wikipedia.org/wiki/Operator_precedence)" in this puzzle.

[billl]

Mick, have you seen the solution to this one? Does the math check out? I have a VERY NEAR solution, but the hand-written alterations (?) have me wondering if there was something amiss along the way. That, combined with the issue of operator precedence. (I actually thought I had the right answer, but I was working in pen and missed the negative sign.)

I have a "solution", but the 36 isn't negative. Maybe I just have to work harder, but I'd like confirmation that there's a solution that checks out, if possible.

Again, the "solution" I have is done without operator precedence... I just went left-to-right, and top-down: multiplication didn't necessarily come before addition. (e.g. 1+2x3=9 is the nature of the solution I found.)

I'll post this "near answer" later if it stumps everyone for too long, and if Mick allows me to do so. That's just a sketchy looking puzzle--but my apologies, of course, if that is your handwriting, Mick.

[prendrelemick]

Ah-ha Well spotted billl. I copied the puzzle straight from an old newspaper - and they have misprinted it - or made a mistake. I reckon that minus sign is wrong too, but it is there in the original.


That'll larn me not to use other peoples puzzles.


I left no instructions to add an extra layer of puzzlement.

[billl]

So you saw the solution, and it didn't add up? Or is it correctly copied from a respectable newspaper?

I may have found a way (well, it isn't as simple as a single 'way') to prove that it is unsolvable, with or without operator precedence--but I'm not a professional mathematician, so... Looking at the mess on my note pad, there's no real reason to be very confident about it at all. Still, the top row and the first two columns produce certain limitations...


here's a tantalizingly close grid for the "using operator precedence" angle (middle row *just misses*):

+1, +8, +8
-1, +1, +13
-10, -13, +2

[prendrelemick]

Sorry billl lets start again.

Ok, I wish I had never started this one, I am a total knobhead at maths.

The grid below is correct . (checked and double checked.)

The boxes along the bottom are totals of the column above them
The boxes down the right are totals of the line to the left of them

Those dashes are minus signs.

All you have to do is fill in the blanks.




http://i85.photobucket.com/albums/k78/prendrelemick/mathsquiz-1.jpg

[MarkBastable]

The problem, for me, is that the number above 21 has to be 21 because 1*x is always x.

And if it is 21, then the number to the right of 21 has to be -1.72, in order to get to -36.

At which point, I rather lose faith in either the puzzle or my understanding of it.

[prendrelemick]

There is a number above the one (or the minus one I should say)
so its something minus 1 times something equals 21

[MarkBastable]

There is a number above the one (or the minus one I should say)
so its something minus 1 times something equals 21

Yeah, which would imply operator precedence and implicit brackets, wouldn't it? I mean, you'd have to have brackets around the (something minus something) in order for that bit to be done before the multiplication.

[prendrelemick]

Start at the end furthest from the total

[billl]

Thanks, Mick! I figured out an angle of attack that I had previously ignored, so I might be looking at it some more tonight.

(I think this is a good one, btw, now that the doubts have been addressed. Maybe it is just really difficult.)

[billl]

Got it!


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

Looks easy all of a sudden. I thought I checked it last night, but I must have left a negative sign in a blank unerased or something.


Here's a crappy attempt to display the whole thing with the symbols:

2 x 4 - 8 = 0
- . - . -
5 - 1 + 6 = 10
x . x . -
3 - 7 x 9 =-36
==============
-9 21 . -7

[prendrelemick]

That's correct billl. I'm such a numpty at maths I didn't know about operator whatsits.

[billl]

I think I may have blown a legitimate concern (operator precedence) way out of proportion, and then cast a stain of suspicion over what turned out to be a challenging and welcome addition to the thread, and I really do apologize Mick. Like Scher, I enjoy getting away from the screen and using pencil and paper for a while, at least some of the time. I just came across some near misses, looked at the hand-written alterations, and figured it wasn't MY fault that I wasn't getting it, or something, so again, I'm sorry about that.

Anyhow, I have no idea where Scher is, but I have another that might be a little fun to work on with pencil and paper. It shouldn't take quite so long, however, as this last one did... With luck, Scher and few more of us will get a crack at it before it gets solved.

If you get the solution in the first few hours, consider sending me a Private Message rather than posting in this thread, and I'll be sure to give you credit as first to solve it (if you are):




The puzzle of the socalled "Poster Variations".

There was recently a (since-deleted) thread on LitNet that explored a topic I dare not revisit, even in vaguest summary. By mentioning that it involved hanging chads and escalatingly ribald personal accusations, I only invite the risk of censorship once more.

However it should be possible to reconstruct the basic structure of the thread by using the clues below. See if you can describe which particular posts were written by which particular poster, and which posts (if any) the particular posts were in response to (a post can only be in response to ONE other post).



Six posts total.
Only one post received more than one reply.
No one replied to the 4th post.
IamNobody was the last person to participate in the thread.
Mark was the last person to participate in the thread.
IamNobody replied to Scher twice.
Scher replied to IamNobody and Mick.
Mark replied to the reply to the OP.

[MarkBastable]

Can you check out the fourth and fifth bulletpoints?

[billl]

Let's call that "word play". Both are true.

[MarkBastable]

Okay - in which case, please check your inbox. I think it covers all the criteria including the wordplay.


Edit: In case you're wondering, I didn't get it quite right. This what comes of trying to do it on a train on your iPhone. I attempted to memorise the criteria and then figure the problem out by drawing with my fingertip in a patch of condensed breath on the train window.

I was travelling, by the way, with a colleague, and I told him I was defining the payment structure for the contract we were travelling to a client site to discuss.

[billl]

Is it about time for a hint?


HINT:
Consider a thread that begins with 4 posts by the following people: Allan, Bob, Bob (again), and then Carl. Who posted first in the thread? Who posted second? Who posted third?

[prendrelemick]

I needed about 8 posts to meet all conditions, but ^ is helpful

[Scheherazade]

Oh, have I been missing all the fun?

I am sorry but have been busier than usual this week. Will give play around with this one today while at work.

[billl]

Well, now that everyone (incl. Scher and Mick) has had a fair chance to consider and take a crack at it, I think it'd be fine if we just posted solutions in the thread as normal. Send me a private message if you want, or if there's a particular question, but otherwise I think using the thread is perfectly fun and fair. Mick just sent me a pretty funny compilation of possible offensive posts, actually...

Anyhow, here's another HINT:

It might be the case that a certain person only posted once, and that he or she was the LAST to participate. However, that doesn't mean he or she posted the last post in the thread... For example, a person that had posted EARLIER than him or her might post again, AFTER he or she did.


EXAMPLE: This is page 39 of this thread. As of this post (number 577), I am the LAST person to participate on this page. However, isn't it also true that I participated before Scheherezade did, on this page...?

Try to deduce who did Post Number One in the puzzle, and then who could have done Post Number Six...

[Scheherazade]

Been trying but not getting anywhere.

Just want to ask whether the same poster can make two consecutive posts as I have been trying to avoid this.

[billl]

Been trying but not getting anywhere.

Just want to ask whether the same poster can make two consecutive posts as I have been trying to avoid this.

They can (but it doesn't happen).



(EDIT: I think you guys can get this one. Again, begin by figuring out who did post number one and who did post number six. It's of course easy for me to say, though... I'm really quite delighted that it's proven such a challenge. But if you feel like you're getting too bored with it to really apply the analysis anymore, and would rather work on something else, please maybe provide some delicate hints... I'm thinking tomorrow or later today, I could give a clue/approach that would make this thing easier, but it might be a dead give-away. What with schedules and other things, for all I know everyone is just getting started on this.

Basically, I don't want to become the dude with the boring puzzle that killed the thread. But I think this thing is doable by you guys.)

[billl]

Another HINT/strategy for solving:

Focus on IamNobody and Scher. I think you should probably be able to determine the order in which their posts occur in the thread.

[Cunninglinguist]

nevermind

[prendrelemick]

How's this (I modified opinions and language for the public domain)

1 Scher- "Chavs are misunderstood children "

2 IAN- replies to (1) "I think they need a jolly good talking to" Replying to scher for the first time

3 Mick replies to (1) "Misunderstood!! String em up, I say" Thus post (1) has 2 replies

4 Mark replies to (2) "**!!@@/#~!!!" Thus Mark replies to the reply of the OP and is the last of the four to participate.

5 Sher replies to (3) "You can't do that!" Thus Sher replies to Mick

6 IAN replies to (5) "He has a point though." Thus she replies to Scher twice and is last to participate.

Ta dar!!

[billl]

Close! However, one of the clues states that Scher replies to Mick and IamNobody. But you most certainly do not have to *totally* revamp your attempt, it really is pretty close.

[prendrelemick]

Ok, brain just about frazzled.


1- Mick is op
2- Scher replies
3- IaN replies to Scher for first time.
4- Mark replies to Scher (2nd reply for 2nd post)
5- Sher replies to IaN
6- Ian replies to Scher for the second time.

[billl]

Yes, that's the solution--thanks for working it through and bringing an end to its tortuous reign. I feel like I should apologize or something, because it was harder than I expected, and it probably made people a little frustrated and crazy. I mean puzzles are supposed to do that, but I can't help but feel guilty taking up five days. To think, my initial fear was that it would be done too soon for everyone to get a shot at it...

I guess there were too many clues, perhaps, and it's a little different kind of logic puzzle than the usual ones with grids.

[prendrelemick]

Don't worry about that, it was up for 5 days but I reckon people only spend a few spare minutes here and there on it.

So, onwards-

Rorden Gamsey, a foul mouthed, scary, celebrity chef, barks an order to the timid Fish Fryer Freddy.

"I need three rare tuna steaks in exactly 3 minutes, or I'll have your guts for garters."

Fish Fryer Freddy knows rare tuna steaks must be fried for one minute on each side. He also knows that his frying pan will only hold 2 steaks at once.

"Yes Chef. No problem Chef" he shouts back.

How does he do it?

[kasie]

He puts two steaks in the pan; after one minute he turns one steak, removes the other steak and keeps it warm and puts in the third steak; after a further minute he removes the first steak, keeps it warm, returns the second steak to the pan to cook the other side and turns the third steak. At the end of the third minute he plates all three rare steaks and shouts 'Ready, Chef.'

[Scheherazade]

Excellent solution, Kasie! :)

Just going back to the previous puzzle (I am not arguing but trying to understand)... Because there were two last posts to the thread, I had assumed that one of those was a direct reply to the OP and the other one was a reply to one of the other posts in the thread. And since Mark was mentioned as having replied to another poster, I, again, assumed that his would be the very last post in the thread and IaN would be the last person to reply to the OP.

I am not sure how the solution given accommodates these. Can someone explain please?



Six posts total.
Only one post received more than one reply.
No one replied to the 4th post.
IamNobody was the last person to participate in the thread.
Mark was the last person to participate in the thread.
IamNobody replied to Scher twice.
Scher replied to IamNobody and Mick.
Mark replied to the reply to the OP.



1- Mick is op
2- Scher replies
3- IaN replies to Scher for first time.
4- Mark replies to Scher (2nd reply for 2nd post)
5- Sher replies to IaN
6- Ian replies to Scher for the second time.

[billl]

Scher, I'm not sure what you are saying when you say that 1) a reply to the OP, and 2) post number 6 being a reply to something in the middle four posts would mean two "last posts" or whatever--I think that the whole "reply" angle is being looked at in a way I hadn't forseen or something, maybe. And I'll admit that the wordplay and spaghetti-type nature of this means I might be missing something pretty interesting.

Here's how Mick's solution works for IamNobody and Mark being "last":
--IamNobody is last because he did the final post (number 6).
--Mark is last because his post comes after everyone else has posted (he was the last one to join the discussion).
In this interpretation/solution, the matter of "replies" doesn't affect who is "last".

AHHHHHHHHHH....
It just occurred to me what you might be getting at!
Are you saying that you had envisioned a pair of "strings of replies"? Like someone replies to someone who had replied to someone, and the last in that string can be considered "the last to participate" in that series of replies? With one string beginning with the OP, and the other string involving the other posts that are not part of the OP's string of replies?

Since that particular pair of clues is built on wordplay, I'd have to say such a solution would be permissable, yeah! (If there is one...)

[Scheherazade]

--Mark is last because his post comes after everyone else has posted (he was the last one to join the discussion).
In this interpretation/solution, the matter of "replies" doesn't affect who is "last".Gotcha!
With one string beginning with the OP, and the other string involving the other posts that are not part of the OP's string of replies?Yeah, something along those lines.
Since that particular pair of clues is built on wordplay, I'd have to say such a solution would be permissable, yeah! (If there is one...)Well, obviously , I could not come with it if it exists! :p Not with 6 posts, at least (I think I managed it with 7, though).

Thanks for the reply, Bill :)

[billl]

No problem! Now I have to see if I can resist the urge to re-check all of the possibilities you already explored using the "strings of replies" interpretation, trying to match your seven.

[prendrelemick]

He puts two steaks in the pan; after one minute he turns one steak, removes the other steak and keeps it warm and puts in the third steak; after a further minute he removes the first steak, keeps it warm, returns the second steak to the pan to cook the other side and turns the third steak. At the end of the third minute he plates all three rare steaks and shouts 'Ready, Chef.'



Correct! Now Kasie your turn, I'm not sure we can accept any more IOU's:hand:

[kasie]

Oh dear, I knew I shouldn't have posted that answer - I am puzzled for a puzzle atm so I will have to think about it for a while.....

OK - I've thunk. I did a quick check and I don't think we've had this one before. If we have, then I apologise.

A farmer goes to market. He buys twenty sheep; he also buys twenty hurdles to make a small temporary fold in which to keep his new acquisitions. He calculates that twenty hurdles will just hold them comfortably. After a swift half at the Dog and Crook (Ramsbottom's Best, of course) he returns to the auction and buys a further twenty sheep, a bargain too good to miss. He then realises that he needs to buy more hurdles as he has doubled the size of his new flock. So - how many more hurdles does he need to buy to double the size of his enclosure? Remember he has already spent more on the sheep than he intended so he doesn't want to spend too much on the hurdles - what's the smallest number he can get away with?

[prendrelemick]

He goes back to the pub, and works out on a beer mat 20 Hurdles arranged 5x5 makes 25 sq Hurdles so twice the sheep will need 50sq Hurdles which is a pen 10x5. As he contemplates the 10 extra Hurdles he will need, he orders a pie and a pint.


As he stares at his round pint glass and nibbles on his Pie an Idea is slowly forming in his head, pie...pie...pi! And he realises he can get away with fewer hurdles by making a round pen. only 5 or 6 extra .

Sorry I can't do the maths (lost my calculator) but the square root of 50 over pi will be the necessary radius, then times 2 and times pi. and that is the number of hurdles (I think.)

[billl]

Imprecise, yes. But flashy!

[Scheherazade]

ESOL / NON-NATIVE ALERT!!!

What are hurdles???

[kasie]

Scher - hurdles are like fencing panels, portable frames with bars or brushwood in them: the farmer can move then around easily to make temporary enclosures. Sorry, it's a bit archaic, it's a very old puzzle, one I dragged back from my long-lost youth.

Mick, that is of course a shepherd's pie, isn't it? Which can be made rectangular, not necessarily round..... Have another pint and re-calculate.

[billl]

7.14 x 7.14

No way is that correct, but I might be just short of the correct answer. Unless Mick is on the right path. Orrrr any other number of reasons for me likely being wrong.

[kasie]

No, the farmer does not have to cut or bend any of the hurdles to make his sheep fold.

[billl]

Mick's math, approximated.

area = pi R*R
50 = pi R*R
R = square root of (50/pi)
R = square root of (15.9)
Circumference = 2 pi R = 2 pi (4.0) = 8 pi = 25.12 = 26 hurdles, so he'd need six more hurdles. And they'd have to be bendable, which Kasie rules out, OR the math gets more difficult and he needs maybe one or two (zero?) extra to make up for the relatively blocky nature of this "circular" pen.

************************************************** **********

Alternately, going back to the 5 x 5 = 25 sq feet being fine for 20 cows, then we can go with 8 x 7 rectangle having 56 sq feet being plenty enough room for 40 cows (and 7 x7 not being quite big enough).

So he'd need 10 more hurdles, if we do an 8 x 7 rectangular enclosure.

Maybe?

[prendrelemick]

Thanks for doing the sums Billl.


Course, if his first pen was round, (or a twenty sided polygon-which is round enough for farmers,) then the assumed 50 square hurdles needed is wrong.

[billl]

Yes, well, we'll work THAT scenario through soon enough.

[kasie]

You're all being far too complicated - this is Kasie, remember, who can't do Maths. This is a theoretical puzzle, they are theoretical sheep, very small theoretical sheep, so don't think animal welfare and how much room a real sheep would need, think x number of units arranged to make a four sided shape the area of which is doubled by the addition of a minimum number of further units.

Please - someone work it out - I'm already packing for New York, I'll be off on Saturday, I'll have to tell you the answer, oh, I wish I hadn't started this one.

[billl]

If it comes down to it, you could PM me or someone else with the answer!

[MarkBastable]

But what's wrong with the answers given so far?

[kasie]

The question was: what's the least number of hurdles he needs to buy to double the area of his enclosure? So far no one has come up with the (very small) number he need buy.

[MarkBastable]

The question was: what's the least number of hurdles he needs to buy to double the area of his enclosure? So far no one has come up with the (very small) number he need buy.

Fair enough. They're wrong.

[billl]

Was the original enclosure a square or rectangle? Was it 5 x 5? Or could it have been 1 x 9, for example? Or a pyramid?

Definitely a square?

[MarkBastable]

Was the original enclosure a square or rectangle?
Definitely a square?

At no point is that made explicit. So I'm starting from the assumption that we can't make that assumption.

[jajdude]

With 20 hurdles (also unfamiliar with this usage of the word) he can make a rectangle that is 7 by 5: area is 35.

XXXXXXX
X..........X
X..........X
X..........X
XXXXXXX

I guess the dots are sheep poop.

With forty sheep he needs an area of 70, 7 by 10.

That's 30 hurdles

XXXXXXXXXX
X................X
X................X
X................X
X................X
X................X
XXXXXXXXXX

And looks like a lot more poop.

Didn't someone already give this answer?

Come to think of it, wouldn't the biggest enclosure be a circle?

How to draw that, and measure the area?

Didn't somebody else also have this idea before me?

Darn it! I'll assume the 20 sheep can fit in a 9 by 3 enclosure:

XXXXXXXXX
X..............X
XXXXXXXXX

Area, who cares?

Double size:

XXXXXXXXX
X..............X
X..............X
XXXXXXXXX

2 more X's

[prendrelemick]

jajdude, you are a genius

[kasie]

Hooray - he's right! The original enclosure was a rectangle 9 x 1 (20 hurdles altogether.) By adding one hurdle at each short end to make a rectangle 9 x 2 (22 hurdles altogether) the farmer has doubled the area of the enclosure at the expenditure of only two further hurdles. (It's hard to draw using only keyboard symbols because the hurdles would join corner to corner and edge to edge.)

Thanks, jj, now I can go off to NY and miss the groans that will be hurled in my direction. See you in a week's time, folks - try to behave while I'm away. Or on the other hand, you could just have fun.

[jajdude]

yeah, stumbled for a while on that one because was thinking too mathematically I guess, which seems to have thrown others off too. Forgetting about "area" seemed to help find the answer.

Will try to think of a new one soon. Don't really have any ideas.

[MarkBastable]

This is an interesting feeling. I'm simultaneously pissed off and impressed.

[billl]

Actually, there is way in which a farmer could have doubled the area of his four-sided enclosure without the addition of any hurdles. If he had begun by enclosing the first 20 sheep with the enclosure's four sides in the shape of a parallelogram with corners set at 45 and 135 degrees, he could double the area of the enclosure simply by shifting the four sides into the shape of a rectangle (ie. setting all the corners at 90 degrees).

[jajdude]

A number with two digits is equal to five times the sum of its digits. If you add 9 to the number, the order of its digits is reversed. What is the number?

[MarkBastable]

Actually, there is way in which a farmer could have doubled the area of his four-sided enclosure without the addition of any hurdles. If he had begun by enclosing the first 20 sheep with the enclosure's four sides in the shape of a parallelogram with corners set at 45 and 135 degrees, he could double the area of the enclosure simply by shifting the four sides into the shape of a rectangle (ie. setting all the corners at 90 degrees).

Except that he didn't know when he bought the hurdles that he'd buy more sheep, so he'd have been unlikely to have arranged the hurdles in such a way that.... Actually I'm not sure we want to get into the possible motives and geometrical inventiveness of the fictional farmer...

[MarkBastable]

a number with two digits is equal to five times the sum of its digits. If you add 9 to the number, the order of its digits is reversed. What is the number?

45...

[jajdude]

yeah, too easy.

[billl]

It's an interesting one, and a little easy is a nice change of pace.

[prendrelemick]

Actually, there is way in which a farmer could have doubled the area of his four-sided enclosure without the addition of any hurdles. If he had begun by enclosing the first 20 sheep with the enclosure's four sides in the shape of a parallelogram with corners set at 45 and 135 degrees, he could double the area of the enclosure simply by shifting the four sides into the shape of a rectangle (ie. setting all the corners at 90 degrees).


Except that he didn't know when he bought the hurdles that he'd buy more sheep, so he'd have been unlikely to have arranged the hurdles in such a way that.... Actually I'm not sure we want to get into the possible motives and geometrical inventiveness of the fictional farmer...


Whereas all the above is true, jajdude was able to find the sheep in amongst all that wool.

[MarkBastable]

Mpe od yjr eomyrt pg pit fodvpnyrmy. ,sfr h;ptopid si,,rt nu yjod dpm pg Uptl/

[prendrelemick]

smf s;; yjr v;pifd yjsy ;pit#f i[pm pit jpidr,

[MarkBastable]

s jptdr" s jptdr"

(smf dp pm)

[MarkBastable]

Shall we say this one's solved, in which case I'll give the solution - or would others like to have a go at it, although Mick'll be taken as the winner?

[billl]

Here's my lame assent to Mick's victory:

http://www.youtube.com/watch?v=y_PZPpWTRTU

[MarkBastable]

Yeah, alright.

Mpe od yjr eomyrt pg pit fodvpnyrmy. ,sfr h;ptopid si,,rt nu yjod dpm pg Uptl/
Now is the winter of our discontent, made glorious summer by this son of York.

Typed using the key to the right of the correct one on a standard QWERTY. Mick responded using the key to the left of the correct one.

[jajdude]

next one Mick?

[prendrelemick]

Its another one of these. BUT this time here are the rules.

Fill the empty squares with numbers that will make the across and down calculations produce the results shown along the bottom and far right. The numbers 1 -9 must appear once only. The calculations should be performed from top to bottom and from left to right (rather than strict mathematical order)


http://i85.photobucket.com/albums/k78/prendrelemick/mathsquiz2.jpg

[jajdude]

may be wrong

[jajdude]

I know it, but will not say.

[MarkBastable]

276
389
415

[MarkBastable]

A quick one. Which is the odd-one-out?


gland
terror
pretty
flash
salads

[Cunninglinguist]

Pretty - it being the only adjective?

[MarkBastable]

Pretty - it being the only adjective?

Nope.

Well, I mean - yeah. But no.

[prendrelemick]

Pretty.

The others have words within words: land error lash lads.

[MarkBastable]

Pretty.

The others have words within words: land error lash lads.

Nope, too.

My feeling is that when respondents come up with a plausible answer, I should not just say they're wrong, but feedback a further clue that demonstrates that it's not that; it's something else. When I said 'yeah' to the first guess, I meant, "Yeah - that's true, but it's not the right answer..." (I won't say whether or not it was the right word.)

So, in response to both the guesses so far....

pure

could be in there, but would not be the odd-one-out...

[prendrelemick]

Remember the great Roy Walker of Catchphrase.


"That's a good answer........But not right!"

[jajdude]

I'll go with "terror."

The others are not as scary.

Still wrong I know.

[billl]

"Salads"

Only one with a repeated vowel in it.

This answer is unbelievably boring and undoubtedly wrong, but it (and jajdude's) might help earn us another example like "pure".

[MarkBastable]

rotor

...would not be the odd-one-out.

[iamnobody]

Gland, as all others can be typed on a single line of a qwerty keyboard.

[MarkBastable]

Gland, as all others can be typed on a single line of a qwerty keyboard.

...Yup.

[iamnobody]

What do the following states (listed alphabetically) have in common?

Alaska
Arizona
California
Connecticut
Hawaii
Illinois
Kentucky
Maine
Minnesota
New Hampshire
New Mexico
Oregon
South Dakota
Texas
Washington

[jajdude]

No idea. They all have 2 or more vowels?

[iamnobody]

Yes they do, but that's not it.

[prendrelemick]

Have they all got a watery boundary.

[billl]

Only if you include rivers, Mick.
(at least there are OTHER states with pretty obvious water boundaries that aren't included in the list, so...)

Gun laws, politics, and order of statehood are also not yielding any obvious reasons for this grouping of states.

[iamnobody]

Hint: If you were looking at a list of all fifty states, it might be obvious.

[MarkBastable]

Hint: If you were looking at a list of all fifty states, it might be obvious.

I am, and it's not.

[billl]

I did too. And I'm an American citizen.


EDIT (I'm looking at an alphabetical list, though. Is that right, or is it the wrong kind of list?)

[iamnobody]

I did too. And I'm an American citizen.


EDIT (I'm looking at an alphabetical list, though. Is that right, or is it the wrong kind of list?)

You're on the right track.
Hint2...It has nothing to do with geography.

[Scheherazade]

2.Alaska
3.Arizona
5.California
7.Connecticut
11.Hawaii
13.Illinois
17.Kentucky
19.Maine
23.Minnesota
29.New Hampshire
31.New Mexico
37.Oregon
41.South Dakota
43.Texas
47.Washington

They are the prime numbered states when alphabetically listed and numbered.

[MarkBastable]

.....wouldn't it be great if that weren't the right answer but just a coincidence?

[iamnobody]

2.Alaska
3.Arizona
5.California
7.Connecticut
11.Hawaii
13.Illinois
17.Kentucky
19.Maine
23.Minnesota
29.New Hampshire
31.New Mexico
37.Oregon
41.South Dakota
43.Texas
47.Washington

They are the prime numbered states when alphabetically listed and numbered.

You got it! Your turn.

[Scheherazade]

.....wouldn't it be great if that weren't the right answer but just a coincidence?We all know there is no such thing as coincidence, Mark.

All right, here is the next one, which I tried to adopt from a Mensa puzzle. Hope it works.


Out of the seven PMs in my Inbox, Mark's is immediately before Papaya's. Bill's is between Mystry's and Mick's. Iamnobody's is immediately after Jajdude and there are two people between Iamnobody's and Mystry's. Mystry's is immediately after Papaya's.


From the top of the list, what is the order of PMs?

[jajdude]

jaj
Iam
Mark
Papaya
Mystry
Bill
Mick

[Scheherazade]

Yes, that is the correct answer. Bill also PMed it to me on Friday so whoever would like to go next :)

(I'm sorry that I did not reply sooner.)

[jajdude]

Go ahead billl, if ya got one.

[billl]

A blacksmith has six sections of chain, each with four links. If it takes him 20 seconds to open and close a link with his tools, what is the minimum amount of time he would need to make one length of chain out of these six sections?

[MarkBastable]

100 seconds? That can't be right - it's too straightforward.

[billl]

Yep, that's right. Too straight-forward. (That is, 100 seconds is not the correct answer.)

[prendrelemick]

80 seconds. he breaks up one of the sections to join the five others.

[billl]

That's it, Mick. (I'd seen this one in a couple books previously)

[prendrelemick]

Oh gawd!


Right, time to scrape the bottom of a really old barrel.


How can 4 be half of 5.

[jajdude]

Reckon it has nothing to do with Roman numerals?

[billl]

Nice work, jajdude. I toyed with Roman numerals for a couple seconds, but not long enough for it to hit me.

[prendrelemick]

An iron clad ivy league question this!

[billl]

I really wish I could do something with 150 A.D.

[MarkBastable]

Wait, wait, wait - where are we now? Jajude thinks he might know. Bill assumes jajude knows. Mick is giving clues that imply jajude's right. Bill's working on variations around the same theme. With all those clues, anyone with a few minutes could work it out, one assumes.

Can we just kill this one - someone give the answer, accept a smattering of applause - and move on?

[prendrelemick]

I really wish I could do something with 150 A.D.



OK, forget the cladding.


Good grief Iv'e got the feeling Jajdude was being iron ic.

[prendrelemick]

[quote=

[jajdude]

So, four = IV and that's half of FIVE (haha, Fe for iron, just got that hint too)

ok

[prendrelemick]

Your turn jajdude.

[jajdude]

I've got one I found tough, required some playing with digits 0 to 9, though two of them are not used. Will need some paper for this one I reckon.

Replace each different letter with a different digit to make

SEND + MORE = MONEY

[billl]

Yeah, good one.

s=9
e=5
n=6
d=7
m=1
o=0
r=8
y=2

_9567
+1085
10653

I'm guessing there aren't any other solutions, but I didn't actually try it out for r=(something besides 8).

[billl]

OK, unless I was dumb enough to triple-check my solution to the previous one incorrectly three straight times, here's the next puzzle, which it is my responsibility to provide.



The Wogged Pursuit of Perfection

In a completely different bubble of the Multi-verse that we lovingly refer to as "M", there is another universe with a similar type of cosmology, ruled over (created, in fact) by a deity known as "Gow". (But he's always been "Pat" to his deity-type-friends.)

Anyhow, Pat was never much for surprises, variety, or anything like that, but he was VERY good at one thing, and that was taking care of business promptly. For Pat, creating an entire planet in a week was a little behind the pace--he managed to create whole planetary systems in that much time. When he finally settled down to business, this is how it went:

For his first solar system, he built one star (sun) with one planet that had one moon. The next week, he built another star (always just the one star per system), but this time he was able to give it two planets, each with one moon. In the next system he built, he put three planets around the star, each planet having its own boring ONE moon orbiting it. And Pat stuck to this formula, oh yes, rigidly, steadily producing solar systems that were gradually (and very predictably) larger than previous ones.

After one year (weeks and years are "Earth years" calculated via atomic clock, for the purposes of this puzzle) Pat stopped. How many planetary bodies (stars, planets, and moons) combined did he have after 52 weeks of building them?

[prendrelemick]

At the half way point on week twenty six he will make 1 sun 26 planets and 26 moons and that is his average output for the year. Thats 53 bodies times 52 weeks, which is 2756. However my maths prowess is legendary - For all the wrong reasons - so I've probably done it wrong.

[billl]

Yes, not quite right. You've shown all the necessary brilliance, and have made a nice presentation of the technique that makes it all so elegant, of course. Just a little bit off. (Lucky for me!)

[jajdude]

2756 + 52 stars ?

[billl]

Yep. (Add first week and last week, then multiply by 26.) Your turn again, jajdude.

[jajdude]

If mick likes, he can find one. I got nothing right now. Can do later with more time to spare.

[jajdude]

OK, a cryptogram:

HSI KZD JST HSCDNE SI NDTJE CH ZPP SZE KIOIPX EHTGGIV HSCDNCDA.

[prendrelemick]

Yep. (Add first week and last week, then multiply by 26.) Your turn again, jajdude.



I mean - why does that work? How can you possibly know to do that?

[MarkBastable]

I mean - why does that work? How can you possibly know to do that?

I don't know whether you're taking the mick, but in case you're not, imagine having to add the numbers from one to five.



* ** *** **** *****

The middle one is the average of the others. Or, to put it another way, you could take two from the last one, and put it on the first one. And one from the fourth one and put it on the second one. You make them all the same because they 'balance' round the middle one.


*** *** *** *** ***

So the sum is the middle number (3) times the number of instances (5).


Same principle with the suns and planets. Figure out how many there are in the middle one and multiply by the number of instances.

[billl]

Except 52 is an even number of elements (with no 'middle' integer--26 is actually the last member of the first half), so one has to take the other (similar) route (summing the extremes and dividing by two to find the average).

[Armel P]

I don't know whether you're taking the mick, but in case you're not, imagine having to add the numbers from one to five.



* ** *** **** *****

The middle one is the average of the others. Or, to put it another way, you could take two from the last one, and put it on the first one. And one from the fourth one and put it on the second one. You make them all the same because they 'balance' round the middle one.


*** *** *** *** ***

So the sum is the middle number (3) times the number of instances (5).


Same principle with the suns and planets. Figure out how many there are in the middle one and multiply by the number of instances.

Of course, 26 would be the middle of 51, not 52.

[Armel P]

Oh. Beaten by billl.

[jajdude]

I mean - why does that work? How can you possibly know to do that?

Once read a story about Gauss, a great mathematician. Teacher wanted to keep kids busy so he/she told them to add numbers 1 to 100. He solved it real fast. I believe he, at maybe 6 years old, took out 100 and the middle number 50. Then it's 99+1 + 98 + 2 + 97 +3 ... or 49 x 100. Or was it 101 x 50, what ever worked. This one had 1 + 2 + ... +52 , or 53 x 26 (twice) as you noted, but forgot to add the stars.

http://www.jimloy.com/algebra/gauss.htm

[billl]

ok, a cryptogram:

Hsi kzd jst hscdne si ndtje ch zpp sze kioipx ehtggiv hscdncda.

zdtdxkr*h?

[prendrelemick]

Thanks for that everyone. I think I understand.- If you add the extremes and divide by two, that must be the average, provided it is a regular progression. But Billl being Billl, was able to jump a step and just times by 26 to get the answer.

[billl]

Whoops! Wow! Finally, a wrapping-up and an end to the avalanche of carelessness (dating back at least to the mix-up (http://www.online-literature.com/forums/showthread.php?t=55937&page=46) about the variable Y in jajdudes puzzle from days ago. 10652!).

EDIT: Correction, I wasn't as careless as I just thought about skipping the division, and multiplying by 26. I think...

[billl]

I have a whopper all loaded up, if you guys really want to deal with it. Definitely requires math, and pencil and paper. I'm taking my time, in case some people are yet to give the cryptogram a fair shot... But raise your objections now, if you must, and I'll try to come up with something else, because you might not like this next one...
:leaving:




EDIT: I've attached the solution to jajdude's cryptogram.

[billl]

Five Barrels

Goran has been assigned the task of ordering a customized trailer for his company. His company makes really big barrels of something (he isn't sure what, he mostly just uses the internet at his desk), and each barrel is 5 feet in diameter. If the trailers can't be wider than 9 feet wide, what is the minimum length the trailer must be to completely accommodate 5 barrels. (The barrels cannot be stacked or lain on their side.)

[jajdude]

Good job bill, will look at your puzzle later. Bit drunk now.

:wave:

[prendrelemick]

There's a coincedence I spent yesterday stacking 5ft round bales into my shed.

[billl]

That might've made for a better set-up than "Some guy has to do something with barrels full of who knows what," but I'm guessing hay is stacked and stored on its side (if drives in the country are sufficient to reveal how that works.)

That sounds great, though, to be outside (and in a shed) dealing with hay. Grass is always greener, maybe, at least some days, but still, sounds nice. Actually, the grass would have lost its green-ness, in this case, and would most certainly be greener over at the other hypothetical spot... Another nice coincidence there, this time in my application of an idiom. Hmph.

Delivering lambs, though--I think I'd have to get pretty good at it before I stopped dreading that particular task.

[prendrelemick]

Its usually wet and cold Billl.

[billl]

Mick got the latest puzzle (The Five Barrels).

I've attached the solution that he PM'ed to me.

[prendrelemick]

Ok, enough of all this logic and maths. Its time for some intuition, a little knowledge and possibly a bit of wiki-ing.


A King fisher is flying over the countryside, when it sees a small pond. Next to the pond is a convenient stone cairn. It lands on the cairn and peers into the water. In the pond is a freshwater fish, that is unfortunetly too large for the bird to catch. My question is, how much stone was needed to build the cairn?

[billl]

Enough stone for the bird to spot a rock perch?

("Perch" has two meanings in this undoubtedly incorrect answer, with one of them being a species of freshwater fish, including something called a "rock perch" which is not actually a perch at all, but is instead a type of small bass.)

[prendrelemick]

Thanks billl, my first clue was going to be that the fish is a perch, and not a red herring. so you're in the right area.

[billl]

24.75 cubic feet of stone?

http://en.wikipedia.org/wiki/Perch_(unit)#Volume

[prendrelemick]

That is correct. The cairn was a perch, and a perch is just under 25 cubic feet of stone. It's not all that archaic either, a few years ago I did a walling job for a large estate and was paid by the perch.

[billl]

The dictionary on my computer mentioned it as a unit to measure length, but didn't have the volume definition so I ended up thinking along different lines. So close...

I'll try and come up with something soon, but anyone else with a good puzzle/problem is welcome to jump in...

[billl]

Ike The Ice Cream Man
Ike the ice cream man has a problem. A bunch of sorority girls just got off at the shuttle bus stop outside his shop, and they are about to come streaming inside his door. Ike can't resist these girls, they get anything they want from him, and he has foolishly left a 750ml bottle of Schnapps that he recently purchased sitting at the side of the counter, in plain view.

The girls love only one thing more than Schnapps, and that's ice cream, and so Ike starts scooping it into cups at the rate of one cup every 10 seconds (the same rate at which the girls are now entering his ice cream shop).

If it takes a girl 60 seconds to finish one cup of ice cream, and 30 seconds to pour and drink a 50ml shot of Schnapps, How long (from the moment he begins scooping) will it be before the sorority girls finish off Ike's Schnapps?

[kasie]

Can I ask, billl - do the girls pour their own schnapps or does Ike stop scooping ice-cream and pour it for them?

[MarkBastable]

Does he keep them supplied with endless cups of ice cream, or does he stop after one each?

[billl]

They pour their own Schnapps, and he supplies as much ice cream as he can (he's still scooping when the bottle's finished).

[prendrelemick]

I can't see a reason why it isn't 7.5 mins.

[billl]

I could have maybe done a better job explaining, the situation is this:

Girls are pouring in at a rate of 1 every 10 seconds, and Ike serves each one of them ice cream at the rate of one cup per 10 seconds (which each of the girls are able to finish in 60 seconds). Those who don't have ice cream will do shots of Schnapps, at a rate of 50ml per 30 seconds, and girls without any ice cream to enjoy will share the bottle. (But any girl with an opportunity to have a cup of ice cream will choose ice cream instead of the Schnapps.)

The girls remain in the shop, eating ice cream as quickly as Ike can serve it to them, and so he is busy with a store filled with sorority girls until the following Friday, when a bunch of mandatory keg parties happen.

[MarkBastable]

So he's okay for the first six girls, but when the seventh girl comes in, he has a problem, because the first girl has finished her ice cream, so he has two girls to give ice cream to, and only ten seconds before another girl's finished and an eighth arrives - and every ten seconds that pass, he has another girl he can't keep occupied with ice cream.

The seventh girl in does a shot of Schnapps in thirty seconds, by which time there are ample girls available to drink further Schnapps.

So from the moment he starts scooping it's 1 minute (while he's holding the first six girls off with ice cream) + 750/50*30secs =

8.5 minutes

The only thing that seems odd to me is that you say it takes thirty seconds "to pour and drink a 50ml shot of Schnapps". I'd'a thought that the next one could be poured as the previous one was being drunk, so what we really care about is how long it takes to pour.

[billl]

The only thing that seems odd to me is that you say it takes thirty seconds "to pour and drink a 50ml shot of Schnapps". I'd'a thought that the next one could be poured as the previous one was being drunk, so what we really care about is how long it takes to pour.

Yes, I had intended to convey the idea that more than one girl could (would) be working on the Schnapp's at the same time. Sorry, again, I'm afraid the correct answer requires a little more jumping through mathematical hoops. (I'll look over my notes again, but I don't think there ends up being a line of people waiting at the Schnapps bottle.)

EDIT: to clarify, more than one girl might be drinking Schnapps at the same time, and each of them would be separately progressing at a pace of 50ml per 30 seconds. There are, as it happens, no cases of more than one girl finishing their drinks at the same time and arguing over the next pour, however (but such a thing could happen in slightly different scenarios).

[MarkBastable]

Yes, I had intended to convey the idea that more than one girl could (would) be working on the Schnapp's at the same time. Sorry, again, I'm afraid the correct answer requires a little more jumping through mathematical hoops. (I'll look over my notes again, but I don't think there ends up being a line of people waiting at the Schnapps bottle.)

EDIT: to clarify, more than one girl might be drinking Schnapps at the same time, and each of them would be separately progressing at a pace of 50ml per 30 seconds. There are, as it happens, no cases of more than one girl finishing their drinks at the same time and arguing over the next pour, however (but such a thing could happen in slightly different scenarios).

So I've obviously gone wrong somewhere, but this was the way it came out for me.

Ike can keep six girls happy with ice cream.

At sixty seconds there'll be seven girls in the store, one of whom'll hit the Schnapps.

At seventy seconds there'll be eight girls in the store - six occupied with ice cream, one with the Schapps and one free. If the Schnapps is free too, she'll hit that - so it depends whether the 'pour' part of 'pour and drink' takes less than thirty seconds. If it takes ten, say, she can hit the Schnapps. If it takes thirty, she has to wait.

By ninety seconds there'll be ten girls in the store - six on ice cream, one starting the second Schnapps (if the thirty-second 'pour and drink' cycles can't overlap) or starting the fourth (if the 'pour' part takes, say, ten seconds), and three hanging about, waiting for some kind of gratification (or two of them could be drinking Schnapps they poured earlier).

But if it's not necessary to know how long the 'pour' bit is, this may be where I've gone wrong - because in my way of working it out, you only need seven girls in the store - six on ice cream (on the 10 and 60 cycle for serve and consume) and one on Schnapps (on the 30 cycle for drink and pour). Any more girls than that are just hanging about. It seems unlikely to me that bill would have set up the problem in this way if he only needed seven girls - so my logic might have gone awry somewhere around there.

[billl]

So I've obviously gone wrong somewhere, but this was the way it came out for me.

Ike can keep six girls happy with ice cream.

At sixty seconds there'll be seven girls in the store, one of whom'll hit the Schnapps.

At seventy seconds there'll be eight girls in the store - six occupied with ice cream, one with the Schapps and one free. If the Schnapps is free too, she'll hit that - so it depends whether the 'pour' part of 'pour and drink' takes less than thirty seconds. If it takes ten, say, she can hit the Schnapps. If it takes thirty, she has to wait.

By ninety seconds there'll be ten girls in the store - six on ice cream, one starting the second Schnapps (if the thirty-second 'pour and drink' cycles can't overlap) or starting the fourth (if the 'pour' part takes, say, ten seconds), and three hanging about, waiting for some kind of gratification (or two of them could be drinking Schnapps they poured earlier).

But if it's not necessary to know how long the 'pour' bit is, this may be where I've gone wrong - because in my way of working it out, you only need seven girls in the store - six on ice cream (on the 10 and 60 cycle for serve and consume) and one on Schnapps (on the 30 cycle for drink and pour). Any more girls than that are just hanging about. It seems unlikely to me that bill would have set up the problem in this way if he only needed seven girls - so my logic might have gone awry somewhere around there.

Yes, you're right--the problem assumes (but doesn't state) that the pouring time is inconsequential. Unfortunately for me, it turns out that my solution involves more than one girl pouring herself Schnapps at the same time on two occasions (during the 50 second period before the bottle is finally emptied).

Anyhow, a bit of a screw-up there, sorry--but you've really saved the day on this one, Mark. You've nailed the issues down perfectly well, and since the replies haven't been exactly pouring in like sorority girls at a late-night ice cream parlour, I'm going to post the solution here in an attachment.

[jajdude]

Too much ice cream and schnapps for me.

Got a good one, Mark?

[MarkBastable]

This one was lifted straight from the 'Net, so it's easy to cheat. I'm going to have only intermittent access to LitNet over the next couple of weeks, so if everyone gets sick of the question, you might all agree to Google it.

--------------------------------------------------------------------


Three men were standing in a row, all facing the same direction, so that there was one in back who could see the two in front of him, one in the middle who could see the guy at the front, and the one in front who could not see either of the other two.

They were shown five hats - three blue and two red.

One hat was placed on each man, without them knowing which, or knowing which two were left over.

First the man in the back was asked if he could deduce what color hat he had on.

"No, I can't," he said.

The man in the middle was asked the same question.

"Nope," he said.

Then man in the front was asked - and he knew what color hat he was wearing.

What was his answer, and why?

[prendrelemick]

I had a beautiful lucid moment just then and worked out it was blue. Trouble is, I can't remember why.

[billl]

I had a beautiful lucid moment just then and worked out it was blue. Trouble is, I can't remember why.

How about this:

The back guy says no, so the guys in front of him can't BOTH be wearing red hats (that would mean only blue ones were left over, so he could deduce "blue").

So the second guy knows that at least one blue must be on either his head or the front guy's head (and maybe both are wearing blue...). If the person in front is wearing blue, then the middle guy can't be sure if he is wearing a blue hat himself--it could be blue OR red. However, if the guy in front of him is wearing a red hat, then he (the guy in the middle) CAN'T be wearing a red hat, because of the situation with the guy in the back (noted above, i.e. he couldn't have seen two red hats without knowing he was wearing blue).

Therefore, since the middle guy can't deduce anything, he must be seeing a blue hat on the front guy. So the front guy deduces that he is wearing a blue hat.

[MarkBastable]

Yep, good. Me and the family are off to NY and the Virgin Islands. Carry on.



How about this:

The back guy says no, so the guys in front of him can't BOTH be wearing red hats (that would mean only blue ones were left over, so he could deduce "blue").

So the second guy knows that at least one blue must be on either his head or the front guy's head (and maybe both are wearing blue...). If the person in front is wearing blue, then the middle guy can't be sure if he is wearing a blue hat himself--it could be blue OR red. However, if the guy in front of him is wearing a red hat, then he (the guy in the middle) CAN'T be wearing a red hat, because of the situation with the guy in the back (noted above, i.e. he couldn't have seen two red hats without knowing he was wearing blue).

Therefore, since the middle guy can't deduce anything, he must be seeing a blue hat on the front guy. So the front guy deduces that he is wearing a blue hat.

[billl]

I would not be at all offended if someone else were to post a new one. (Actually, Mick got the correct answer... so he's equally on the hook.)

I will put something up eventually though, if nobody has thought of any other good ones by then.

[prendrelemick]

How about if Mark leaves Heathrow at 06.00 oclock and lands in New York at 08.00. How long is he in the air?

[MarkBastable]

How about if Mark leaves Heathrow at 06.00 oclock and lands in New York at 08.00. How long is he in the air?

It depends how many people under five you're travelling with. A few years back, when Nell and Grace were four and two respectively, I seem to remember being on a flight for several days.

[MarkBastable]

Also taken from the Net, and a variation on the previous one...

Three women - one of whom we'll call Intelligentsia, because she has a talent for logic problems, and is therefore an absolute blast at extended wine-bar lunches - are sitting in a circle. They are blindfolded and told that a spot of either red or white will be put on each of their foreheads.

A red spot is then put on each of their foreheads.

The blindfolds are removed and they are told to raise their hand if they can see any red spots. All three raise their hands.

They are told to lower their hands if they know what colour spot they have on their own forehead.

Intelligentsia ponders for a moment, and then lowers her hand.

What has she figured out that the other two haven't?

[billl]

uh, uh..

[prendrelemick]

All those wine bar lunches paid off, she can smell the Pinot Noir on her forehead.

[MarkBastable]

Because precision of language is so important in these things, I've edited the problem slightly.


...just as a clue - the other two women are pretty bright too. But not as bright as Intelligentsia.

[billl]

Oh, one of the others has a white spot on their head. So the OTHER (3rd) woman must've seen a red spot on Intelligentsia's head.

[MarkBastable]

Oh, one of the others has a white spot on their head. So the OTHER (3rd) woman must've seen a red spot on Intelligentsia's head.

No, they all have red spots.

[billl]

It has something to do with the plural "spots"?

[billl]

How about if Mark leaves Heathrow at 06.00 oclock and lands in New York at 08.00. How long is he in the air?

Seven hours?

[MarkBastable]

It has something to do with the plural "spots"?

Nope - there's nothing tricksy about it. The problem is just as it appears.

[billl]

But if Intelligentsia can see a red spot on each of the other women's heads, then it doesn't matter if she has a red or white spot--the other women would each see (at least the one) red dot. How could she be sure she didn't have a white dot?

Is it because the others would've known that THEY would both have a red dot if she had a white dot and didn't announce that she was sure that she had a red dot? And so the pause means no one can be sure, which means: three red dots...?

I see, then! So all three are smart enough to figure it out, if the given the visible dots are sufficient--but she is the first to figure out that she can, on account of their uncertainty, figure it out even when the visible dots had left some ambiguity...

[MarkBastable]

If the others are logical, Intelligentsia can be completely certain she has a red spot - unambiguously.

But you're damn close (in fact, you might be right, but I'm not certain about the phrasing).






But if Intelligentsia can see a red spot on each of the other women's heads, then it doesn't matter if she has a red or white spot--the other women would each see (at least the one) red dot. How could she be sure she didn't have a white dot?

Is it because the others would've known that THEY would both have a red dot if she had a white dot and didn't announce that she was sure that she had a red dot? And so the pause means no one can be sure, which means: three red dots...?

I see, then! So all three are smart enough to figure it out, if the given the visible dots are sufficient--but she is the first to figure out that she can, on account of their uncertainty, figure it out even when the visible dots had left some ambiguity...

[prendrelemick]

Seven hours?



Is it? Crikey!

[prendrelemick]

OK, this is work in progress

If Intelligensia had a white dot, then the other two would know they had red dots, because all three had seen a red dot. (so at least two dots had to be red) As they didn't know, Intellegensia's had to be red.

Does that make sense?

[billl]

Actually, I just re-read what I wrote late last night, and it is really, really a mess. In particular, Intelligentsia's hypothetical behavior in the middle portion makes no sense. And near the end, where I say "ambiguous", I was just trying to convey the "As they didn't know" phrase that Mick uses in his solution.

[jajdude]

Can we go back to sheep and farmers? At least I got that one.

[billl]

I still think my answer in post 728 has something to it: Intelligentsia would have raised her hand (immediately) if she had seen that the other women had one white and one red dot between them. Since the woman with a red dot had acknowledged seeing red, that would mean that Intelligentsia would have to have a red dot as well.

The thing that threw me off was my misreading of the problem. It specifically states that all three have red dots. (I'm wondering if there might be a series of "Dotted Head" puzzles, with slight variations like this...)

[billl]

Can we go back to sheep and farmers? At least I got that one.

I think we are all free to come up with one now, but I might take a day or two, unless I luckily stumble upon one quickly. I will keep your preference in mind, though.

[prendrelemick]

I still think my answer in post 728 has something to it: Intelligentsia would have raised her hand (immediately) if she had seen that the other women had one white and one red dot between them. Since the woman with a red dot had acknowledged seeing red, that would mean that Intelligentsia would have to have a red dot as well.

The thing that threw me off was my misreading of the problem. It specifically states that all three have red dots. (I'm wondering if there might be a series of "Dotted Head" puzzles, with slight variations like this...)


But then they would all know what dot they had - I think.

[billl]

But then they would all know what dot they had - I think.

Good point (I love this kind of puzzle). Still, Intelligentsia would get the win (first lowered hand?) because of her noted mental superiority.

(I think the one woman with a white dot wouldn't know *at first*, but the other red dot woman would know, same as Intelligentsia. Of course, after judging both of the red dot women's reactions, I guess the white-dotted woman could deduce her own white dot, much the same as Intelligentsia deduces her red dot in the original puzzle based on the reaction of the others... That's the interesting twist with some of these angles--it depends on judging the reactions of the others).

[prendrelemick]

Ok, time for a quickie.


I remember an interview given by Muhammud Ali before the "Rumble in the Jungle" fight. He said something along the lines of :- "Why I'm so fast, when I turn out the light by the bedroom door, I can be across the room and into bed before it goes dark." Typical Ali boastfull humour I thought. But then I realized he could actually do it.
How.

[MarkBastable]

Luminous wallpaper.


Or, with a big fight scheduled for the following day, he gets a really early night.

[prendrelemick]

Yes, one of those answers is right.:smilewinkgrin:


For jajdude...

http://homepages.paradise.net.nz/wmg/Sheep.html

[jajdude]

Yes, one of those answers is right.:smilewinkgrin:


For jajdude...

http://homepages.paradise.net.nz/wmg/Sheep.html

baad joke

[MarkBastable]

12-26-21-15 29-18-25-26-15 15-26-14-19-17-25-26 16-19-36-23-13-14 12-18-36-23-19-12 _-_-_-_-_-_?

[prendrelemick]

35-14-13-21-28-22

[Delta40]

Ok, time for a quickie.


I remember an interview given by Muhammud Ali before the "Rumble in the Jungle" fight. He said something along the lines of :- "Why I'm so fast, when I turn out the light by the bedroom door, I can be across the room and into bed before it goes dark." Typical Ali boastfull humour I thought. But then I realized he could actually do it.
How.

He goes to bed before the sun goes down.

[MarkBastable]

35-14-13-21-28-22

I thought that would take longer. Have we used that code before?