zex20913
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Oh man, I must have been stupid this morning, or something. I meant, of course ONE Pi.

But now, the answer is easy. 19. The flip side of it is 61, and there are only 60 minutes in an hour.

That is, unless there's no colon between the hours and minutes.

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Mattcrampy
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Am I allowed to feel gypped?

Matt

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What do you call an elephant at the North Pole?

zex20913
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Feel away

If you want it, I'll hand it off to you.

I was waiting for confirmation, but I just realized that I shouldn't. Do you want it?


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Mattcrampy
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Nah, I didn't get the right answer. It's yours.

Matt

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levelthirteen
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zex20913, back to you.

zex20913
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Alright. Sorry for the delay there folks.

A doctor gives you five pills, and tells you to take one every hour and a half. How long do they last?

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Mattcrampy
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Six hours. You take one at 0 hours, then another at 1.5 hours, then another at 3 hours, then another at 4.5 hours, then the last one at 6 hours.

That right?

Matt

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What do you call an elephant at the North Pole?

zex20913
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Yup. Your turn.

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Mattcrampy
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I had this good one, but I lost it. Shame.

But this one's good as well:

I have an unmarked scale with 40 weights, from 1 kilogram to 40 kilograms. I only ever want to weigh things to the nearest kilogram (I'm not that concerned about accuracy) and my poor scale can't take more weight than about 40 kilos, so I won't be weighing anything heavier than that.

I don't need all these weights. How many weights do I need to weigh anything up to 40 kilograms, and what are they?

Matt

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eytanz
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You need 5 weights - 1, 2, 4, 8 and 32 kilograms.

Sell the remaining 35 weights, and use the money to trade in your scales for stronger ones, so that you can measure up to 63 kilos.

[Edited by eytanz on 01-20-2004 at 06:24 AM GMT]

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Schik
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I can beat 5 - use weights on both sides.

1, 3, 9, 27

1 = 1
2 = 3-1
3 = 3
4 = 3+1
5 = 9-3-1
6 = 9-3
7 = 9-3+1
8 = 9-1
9 = 9
10 = 9+1
11 = 9+3-1
12 = 9+3
13 = 9+3+1
14 = 27-9-3-1
...
40 = 27+9+3+1

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eytanz
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Schik wrote:
I can beat 5 - use weights on both sides.

1, 3, 9, 27

Good

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Mattcrampy
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Schik has it.

Daym. I'm fishin' out my other one.

Matt

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Schik
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There are four towns that lie at the corners of a 10 mile square. Let's call them A, B, C, and D. The DOT wants to build roads linking them all together. Since these towns are very poor, they can only afford to build the shortest total length of roads possible. You must be able to get from any town to any other town on these new roads.

How many miles of roads must be built, and how would they be arranged?


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zex20913
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Must it be a direct route from town to town? (I.E, A-C can't go through B?)

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Schik
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zex20913 wrote:
Must it be a direct route from town to town? (I.E, A-C can't go through B?)

It does not have to be direct. As long as you can get from any town to any town using some road or roads, it's okay.

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The_Red_Hawk
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Two diagonal roads across, meeting in the middle.

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Snapping, turning, rising, swirling,
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The red hawk's dance of death.

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Watcher
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The_Red_Hawk wrote:
Two diagonal roads across, meeting in the middle.

I know this puzzle, and I can tell you that it can be done better.

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The_Red_Hawk
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Let's see..... two diagonal roads would be the square root of 200, or 14.142..........

Unless you can only use a one connecting road and a bit, it's impossible.

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Slashing, whirling, diving, twirling,
Snapping, turning, rising, swirling,
Screeching, flipping, gliding, sliding,
The red hawk's dance of death.

.....the king of the skies.....

Schik
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The_Red_Hawk wrote:
Let's see..... two diagonal roads would be the square root of 200, or 14.142..........

Unless you can only use a one connecting road and a bit, it's impossible.

It's not impossible. Your solution is not bad, but it's not the best.

And you're forgetting to multiply your 14 by 2, since there are two roads.



[Edited by Schik on 01-20-2004 at 07:12 PM GMT]

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agaricus5
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I think I have it.

The roads must be arranged like this:

Label the towns A, B, C and D clockwise round the square. Extend one road from A and another from B into the square in such a way that they meet at a point E where AE = BE, angles ABE and BAE = 30 and angle AEB = 120. Do the same to towns C and D so that the roads coming out from them meet at a point F where CF = DF, angles CDF and DCF = 30 and angle CFD = 120. Now join points E and F together.

The total length of road = 4(10/(3^(1/2)))+ (10 - 10/(3^(1/2)))

Which comes to about 27.321 miles.

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Watcher
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agaricus5 wrote:
The total length of road = 4(10/(3^(1/2)))+ (10 - 10/(3^(1/2)))

Just for your information, that can be simplified to 10 + 10*3^(1/2).

This is the solution I had in mind. But you still have to prove that it is the best solution. (I think... but maybe Schik would disagree.)

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Schik
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That's the answer I have, and I won't require a proof. agaricus5 is up!

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Oneiromancer
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You forgot to account for the additional cost due to the construction workers screwing things up because it's such a precise and unusual design. It would actually end up being cheaper to do the X layout because they'd be more likely to get it right.

Well, okay, it's not that kind of puzzle, but I just couldn't help myself.

Game on,

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agaricus5
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Yay!

I do have a puzzle this time.

However, it may be too mathematical for some people's tastes, so apologies in advance if I break the rule about over-complicated mathematics.

OK.

I have 100 metal balls. They are all perfect spheres with a diameter of 2 cm, with no blemishes or bumps on them and are all identical. I am packaging them for a friend who is collecting them, so they must stay whole and intact (undamaged) for him. What is the volume of the smallest box that I could fit them all into? (The box must be a cuboidal shape and assume that the thickness of the sides of it are negligible)

[Edited by agaricus5 on 01-20-2004 at 09:34 PM GMT]

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Oneiromancer
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Well, that sounds like it's taken straight out of my Solid State Physics book for an HCP lattice or something. If someone else complains I'll ask you to pick a new one, but until then I'll let it go.

Game on,

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agaricus5
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Oops.

I didn't realise.

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Schik
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I'd melt all the balls down, and fit them in a box with a volume of 400*PI/3 * R^3, where R is the radius of the balls.

I'm pretty sure that's not what you're looking for though

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agaricus5
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Unfortunately, Schik, it's incorrect but certainly a good answer . Let me correct the puzzle.

The balls of diameter 2 cm are being packaged for a friend, who wants them whole and intact (undamaged) for some sort of collection.

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Scott
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The smallest cubeoid that is 100 or more is 125. That gives 5 balls across and 5 up with some space left at the top. 5x2 is 10 cm a side. The volume of that cube would be 1000.