Daily puzzles/problems.

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

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/k7...athsquiz-1.jpg

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.

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

Quote:

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Originally Posted by prendrelemick

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

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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.

Start at the end furthest from the total

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.)

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

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

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.

Can you check out the fourth and fifth bulletpoints?

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

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.

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?

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

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.

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...

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.

Quote:

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Originally Posted by 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.

---

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.)

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.

nevermind

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!!

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.

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.

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.

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?

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.'

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?

Quote:

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Originally Posted by billl

• 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.

---

Quote:

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Originally Posted by prendrelemick

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.

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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...)

Quote:

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Originally Posted by billl

--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!

Quote:

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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.

Quote:

---

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 :)

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.

Quote:

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Originally Posted by 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.'

---

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

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?

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.)

Imprecise, yes. But flashy!

ESOL / NON-NATIVE ALERT!!!

What are hurdles???

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.

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.

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

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?