TripleM
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Hey, I want to know the answer to your last one

UrAvgAzn
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First post in this thread, hope I'm doing it right.

'Twas a dark, and stormy night. There were two owners of a mansion, one already dead, and one just murdered. There are five, yes, five suspects. The suspects are the Butcher, the Maid, the Guard, the Gardener, and the Dog. The Butcher said he was too busy because he was sharpening his meat knives. The Maid said she was making the Third Guest Bedroom (yes, the one with the creepy spiderwebs), the Guard said he was outside guarding the gate, the Gardener said she was watering the plants, the Dog err... well... said 'Woof'. We roughly translated that as 'I was stealing eating the Cat's food'. Who killed the Second Owner?

Keep posting,

P.S. Are we supposed to mod up the correct guesser? If so, by how much?

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NiroZ
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UrAvgAzn wrote:
First post in this thread, hope I'm doing it right.

'Twas a dark, and stormy night. There were two owners of a mansion, one already dead, and one just murdered. There are five, yes, five suspects. The suspects are the Butcher, the Maid, the Guard, the Gardener, and the Dog. The Butcher said he was too busy because he was sharpening his meat knives. The Maid said she was making the Third Guest Bedroom (yes, the one with the creepy spiderwebs), the Guard said he was outside guarding the gate, the Gardener said she was watering the plants, the Dog err... well... said 'Woof'. We roughly translated that as 'I was stealing eating the Cat's food'. Who killed the Second Owner?

Keep posting,

P.S. Are we supposed to mod up the correct guesser? If so, by how much?

I would say that the dead owner killed the other owner before the first owner died. Why else would you mention him?

Niccus
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Ooh, I know the answer... But then, I don't have any puzzles handy. Shame.

zex20913
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Hmm...I'd guess the First Owner.

"one already dead, the second one just murdered." "second" doesn't necessarily imply "Second Owner", and when I take "murdered" as a verb...well, it's all pretty clear who the murderer is.

The cat.

J/K. First owner.

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Maurog
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I'd guess the Maid, because why making the guest room if there aren't any guests (if they were, they'd be suspects too)?

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Schik
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I'll guess the gardener. If we assume the garden is outside, then why is she watering the plants if it's "dark and stormy"?

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Jason
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There was another puzzle like this but it was a round house and the maids were "Cleaning the corners" which don't exist.

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Mattcrampy
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Well, if the butcher's knives are sharp, the cobwebs are broken in the third Guest bedroom and the plants are wet then all three servants have an alibi. The guard probably doesn't. The dog also doesn't, and the original owner is indeed a suspect unless his body's present (the ultimate alibi: I couldn't have done it, I've been dead for years!)

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Fafnir
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TripleM wrote:
Hey, I want to know the answer to your last one

Christmas = 25 Dec(ember)
Hallowe'en = 31 Oct(ober)

"Dec" can be read as shorthand for "Decimal" - base ten. "Oct" can be read as shorthand for "Octal" - base eight. 31 in octal is equal to 8 * 3 + 1 = 25 in decimal. Hence, Christmas is mathematically equal to Hallowe'en.

Another obscure joke in bases for Hitchhiker's Guide fans: six times nine actually does equal 42 - in base thirteen.
OK, so Douglas Adams said it wasn't intentional. But it's still funny.

I'd be kind of interested in that pictorial explanation of the chain rule, myself - I don't see how to prove it intuitively.

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TripleM
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Fafnir wrote:
I'd be kind of interested in that pictorial explanation of the chain rule, myself - I don't see how to prove it intuitively.

That was the 'last one' I was referring to

zex20913
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Fafnir wrote:

Another obscure joke in bases for Hitchhiker's Guide fans: six times nine actually does equal 42 - in base thirteen.
OK, so Douglas Adams said it wasn't intentional. But it's still funny.

How unlucky.

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UrAvgAzn
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Schik wrote:
I'll guess the gardener. If we assume the garden is outside, then why is she watering the plants if it's "dark and stormy"?

Good job Schik! Your turn. So I'll just +1 that post since it's already +1ed. Okay, so did I do it right?

Keep posting,

____________________________
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Such grief, such joy, to live at all."
- T.A. Barron, The Lost Years of Merlin

Schik
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Yay! Yeah, you did fine. We generally +1 the winners here, so you didn't even need to mod my post. But I won't complain.

So, new puzzle:

You have two pegs, a string, and a picture frame. The pegs are in a wall, and the string has its ends attached to the two top corners of the picture frame. Your task is to hang the frame on the pegs in such a way that removing either of the two pegs from the wall will make the frame crash to the floor.

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Chaco
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Define "crash".

If it just means "make contact with the floor" I think I can solve this one if I have a veeeeery long piece of string.

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stigant
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Can you hang the picutre upside down?

+---------+
|         |
| _______ |
|/o     o\|
+---------+

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Progress Quest Progress

coppro
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Schik wrote:
Yay! Yeah, you did fine. We generally +1 the winners here, so you didn't even need to mod my post. But I won't complain.

So, new puzzle:

You have two pegs, a string, and a picture frame. The pegs are in a wall, and the string has its ends attached to the two top corners of the picture frame. Your task is to hang the frame on the pegs in such a way that removing either of the two pegs from the wall will make the frame crash to the floor.

Simply stick the frame on top of the pegs, and wrap the string around the frame so that it doesn't catch.

TripleM
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In case anyone was wondering, this isn't a trick question Theres a very nice solution (well, infinitely many, I guess).

Schik
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Yeah, what TripleM said. The string will be in contact with the pegs when it's hung from both of them. The string is not so long that taking out one peg will make the frame reach the floor. And the frame is not upside down. I must say I like stigant's solution though - I didn't see that one coming.

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Maurog
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Hmm, maybe the string is just looped on top and caught between the two pegs? The gap between the two pegs is not wide enough to let the loop come trough, so it should be strong enough to hold the picture.

  ___
  \ /
  OXO
  / \
 /   \

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NiroZ
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Maurog wrote:
Hmm, maybe the string is just looped on top and caught between the two pegs? The gap between the two pegs is not wide enough to let the loop come trough, so it should be strong enough to hold the picture.

  ___
  \ /
  OXO
  / \
 /   \

No, tie a knot in the string, and have it placed between two pegs in a way that it doesnt' fall through.

michthro
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File: twopegs.PNG (3.9 KB)
Downloaded 278 times.
License: Public Domain

Something like this:

---

[Last edited by michthro at 11-18-2006 11:59 AM]

NiroZ
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File: 1 peg.PNG (3.6 KB)
Downloaded 72 times.
License: Public Domain

< So it turns out like this?

I know, I thought of that too.

---

[Last edited by NiroZ at 11-18-2006 12:14 PM]

zex20913
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michthro wrote:
Something like this:

Click here to view the secret text

×

This really looks like an alien looking out of a cardboard box.

But if I interpret your picture correctly, directions:

left corner, over peg1, under peg2, back over peg2 and peg1, under peg1, over peg2, right corner.

Then it should be correct.

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michthro
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File: twopegs2.PNG (3.6 KB)
Downloaded 258 times.
License: Public Domain

zex20913 wrote:
This really looks like an alien looking out of a cardboard box.

It does, doesn't it? Sorry about the poor quality of the drawing.

But if I interpret your picture correctly, directions:

left corner, over peg1, under peg2, back over peg2 and peg1, under peg1, over peg2, right corner.

That's it, yes.

NiroZ wrote:
< So it turns out like this?

No, if you remove a peg, it turns out like this:

---

[Last edited by michthro at 11-18-2006 01:53 PM]

Schik
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Yep, michthro got it. See, no tricks. You're up!

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michthro
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Ok, here's one of my all-time favourites:

Given a rectangle R divided into smaller rectangles, prove that if every smaller rectangle has a side of integral length, then R has a side of integral length.

Fafnir
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Proof by induction:

Start off with a single rectangle A with one integral side X and one non-integral side Y. Add another rectangle B to it, also with one integral side and one non-integral side, to form a third rectangle A + B. Trivially, B must have the same side length as A on the side upon which it is placed if A + B is to be a rectangle.

If B is placed onto side X (i.e. one of B's sides must be of X's length), then A + B must have one side of length X - i.e. one integral side. So A + B has an integral side.

If, on the other hand, B is placed onto side Y, then one of B's sides must be of length Y - i.e. non-integral. Therefore, B's integral side must be directly aligned with X. So one of A + B's sides is given by the length of B's integral side plus the length of A's integral side, and so must be integral itself. So A + B has an integral side.

We see now that in either case, adding a rectangle with one integral side to another rectangle with one integral side in such a manner as to produce a third rectangle will always produce a third rectangle with one integral side. This process can be repeated ad lib to produce any type of composite rectangle (i.e. adding C onto A + B, then D onto A + B + C, and so on to arbitrary complexity). Therefore, since any composite rectangle of the type described can be produced by this method, and every rectangle produced by this method has one integral side, every composite rectangle of the type described has one integral side.

I can see why it's one of your favourites - quite a nice solution! Sorry the proof's so long - mathsy language is the easiest type to use when working this sort of thing out, and it's too late for me to translate to English...

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TripleM
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File: rect.JPG (2.6 KB)
Downloaded 206 times.
License: Public Domain

I'm with you until..

Fafnir wrote:
Therefore, since any composite rectangle of the type described can be produced by this method

here.
What about something like this:

---

[Last edited by TripleM at 11-18-2006 11:16 PM]

michthro
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Yep, TripleM's right. Nice try, though. Another thing: Both side-lengths of A can be integral. That's a trivial case, but just to be sure we're talking about the same problem.