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Author Topic: A Riddle
Tante Shvester
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A sundial is the timepiece with the fewest moving parts. What is the timepiece with the most moving parts?
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Strider
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define timepiece.
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Elmer's Glue
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Hourglass.
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Tante Shvester
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A device used to measure the passage of time, I guess. Does this suffice?
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EmpSquared
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The Earth?
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Strider
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damn, i was going to go with that one.

the universe?

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Lyrhawn
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A tree.

(I think hourglass is the best answer so far)

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Elmer's Glue
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This watch.
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Tante Shvester
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Elmer's got it!

Good job. [Big Grin]


Edit:
Clarification --
The correct answer is hourglass.

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Elmer's Glue
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[Cool]
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Tante Shvester
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Do you have any good riddles for me?
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Tatiana
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That's a good riddle. I think one of those water clocks might have even more moving parts, if you think of the water as a very fine sand of molecules. [Smile]
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Humean316
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I have one:

Three grainsilos have the following capacity:

A 8000 kilos
B 5000 kilos
C 3000 kilos

A is full, B and C are empty.

Can you without weighing put 4000 kilos in silo A and 4000 kilos in silo B?

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EmpSquared
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Put 3000 in C
Transfer that 3000 to B
Fill C up with the rest from A so that:

A=2000
B=3000
C=3000

Fill up B to the top with C, so the remainder of C is 1000.

A=2000
B=5000
C=1000

Dump C into A.

A=3000
B=5000
C=0

Fill up C with B.

A=3000
B=2000
C=3000

Fill up A with C.

A=6000
B=2000
C=0

Fill up C with B.

A=6000
B=0
C=2000

Fill up B with A.

A=1000
B=5000
C=2000

Fill up C with B.

A=1000
B=4000
C=3000

Fill up A with C.

A=4000
B=4000
C=0

Voila. Was that too many steps?

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Humean316
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Well, when you dump C into A and get 3,5,0 consecutively, you could have done that first by dumping A into B, but essentially you are correct. The website where I got the riddle solved it another way, so I'll make that a challenge to someone else to solve it another way.
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TomDavidson
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A: 8000
B: 0
C: 0

A: 5000
B: 0
C: 3000

A: 5000
B: 3000
C: 0

A: 2000
B: 3000
C: 3000

A: 2000
B: 5000
C: 1000

A: 7000
B: 0
C: 1000

A: 7000
B: 1000
C: 0

A: 4000
B: 1000
C: 3000

A: 4000
B: 4000
C: 0

A: 0
B: 4000
C: 4000

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Javert
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quote:
Originally posted by TomDavidson:

A: 0
B: 4000
C: 4000

Impressive.

How'd you get 4000 in the silo that could only hold 3000? [Wink]

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scifibum
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quote:
Originally posted by Humean316:
I have one:

Three grainsilos have the following capacity:

A 8000 kilos
B 5000 kilos
C 3000 kilos

A is full, B and C are empty.

Can you without weighing put 4000 kilos in silo A and 4000 kilos in silo B?

A 8000
B 0
C 0

A 3000
B 5000 <--count the grains as you add them, total is N
C 0

Move N/5 back into A

No weighing. [Smile]

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Strider
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I'll do my own alternate method now!

A. 8
B. 0
C. 0

A. 3
B. 5
C. 0

A. 3
B. 2
C. 3

A. 6
B. 2
C. 0

A. 6
B. 0
C. 2

A. 1
B. 5
C. 2

A. 1
B. 4
C. 3

A. 4
B. 4
C. 0

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Paul Goldner
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My alternative solution.

Chop A in half, slip a sheet of steel between the halves, flip the top half over, call it B.

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Strider
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I think I win for least amount of steps, and least amount of insanity.
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scifibum
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Can I win for most amount of insanity?
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Humean316
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quote:
Originally posted by scifibum:
Can I win for most amount of insanity?

Yes, yes you can.

But Strider wins the thread because that's how the people who came up with the puzzle solved it. Well done!

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Noemon
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Oh, oh, I've got a great riddle. It's really obscure, too!

What walks on four legs in the morning, two legs at noon, and three legs in the evening?

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neo-dragon
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A human, although that isn't so much a riddle as an exercise in metaphor.
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ClaudiaTherese
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A fish!
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ClaudiaTherese
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No ... a table! I know it's something about a table.
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T:man
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stand
I

Edit: oopsie

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The Rabbit
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Here's a great puzzle.

You have 10 bombs and you know that one of them is a dud. The dud has different weight (mass) than the bombs but you don't know if it weigh more or less. You have a balance, the kind where you can put things on both sides and see if they weigh the same.

It is possible to identify the dud in three weighings. How do you do it?

There are no standard counter weights available. The only thing you can put on the balance is the bombs.

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scifibum
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Never mind.
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BlackBlade
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Douse all the bombs with some sort of flame accelerant, light up all the bombs, pick up the one that does not explode. Place all the casings and debris on the scale for somebody else to pick up.
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andi330
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quote:
Originally posted by Noemon:
Oh, oh, I've got a great riddle. It's really obscure, too!

What walks on four legs in the morning, two legs at noon, and three legs in the evening?

That is totally from The Hobbit. It's a human.
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BlackBlade
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quote:
Originally posted by andi330:
quote:
Originally posted by Noemon:
Oh, oh, I've got a great riddle. It's really obscure, too!

What walks on four legs in the morning, two legs at noon, and three legs in the evening?

That is totally from The Hobbit. It's a human.
It's totally not applicable anymore either. Now adays many of the infirm have walkers and wheelchairs. I don't see a whole lot of canes these days.
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ClaudiaTherese
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quote:
Originally posted by Noemon:
Oh, oh, I've got a great riddle. It's really obscure, too!

Sarcasmometer reading: 9.3 on the 10-scale

( [Wink] )

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Flaming Toad on a Stick
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Rabbit:
Bombs numbered 1-10
1-3vs4-6(M1)
If even, 1-3vs7-9(M2). If even, 10 is your dud. If (M2) not even, you now know that dud is in 7-9 group, and whether it's lighter or heavier. 7vs9 (M3), if even, 8 is dud. If uneven, the dud will be apparent based on (M2), either 7 or 9.

If (M1) uneven, 1-3vs7-9(M2). If even, dud is in 4-6 group. If uneven, dud is in 1-3 group. In either case, the dud's discrepancy will be known.

In short:
1-3vs4-6(M1)
1-3vs7-9(M2)
E=even, U=uneven
EE, 10.
EU, 7,8 or 9, discrepancy known.
UE, 4, 5 or 6, discrepancy known.
UU, 1, 2 or 3, discrepancy known.
(M3), if necessary, is any two bombs from the group containing the dud. The dud will be obvious in either case.

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The Rabbit
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Flaming Toad: You have a problem.

(M3) Won't work as you have it written. If you select two of the possibly bad bombs and they come out even -- then its the third bomb. But if they come out uneven, all you know is that one of those two is bad. You need one more detail.

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scholarette
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But from the first two weighings, you know if the bomb is too heavy or too light.
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Threads
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I think Flaming Toad's method provides all the information we need. Here's another way of looking at it.

[1,2,3] vs. [4,5,6]
[4,5,6] vs. [7,8,9]

If the groups are all equal then 10 is the dud.

If they aren't then one of them either weighs less than or more than the other two and the dud therefore weighs less than or more than the non-duds respectively (without loss of generality we have A=B and (B<C or B>C), if B<C then the dud weighs more than non-duds and if B>C then the dud weighs less). Toad's M3 works in both cases you listed. If the two bombs picked weigh the same then the dud is the third. If they don't then choose the lighter one if the dud is lighter and the greater one if the dud is greater.

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kmbboots
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quote:
Originally posted by andi330:
quote:
Originally posted by Noemon:
Oh, oh, I've got a great riddle. It's really obscure, too!

What walks on four legs in the morning, two legs at noon, and three legs in the evening?

That is totally from The Hobbit. It's a human.
That is totally considerably older than Tolkien.
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The Rabbit
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quote:
Originally posted by scholarette:
But from the first two weighings, you know if the bomb is too heavy or too light.

That's what I was looking for.


Now the harder puzzle. Same as above but you have 12 bombs.

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vonk
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Re: the original riddle, I was going to guess the Solar System. Does that not have more parts than an hourglass, or is it not an accurate enough timepiece?
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anti_maven
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To continue the theme:

"What have I got in my pocket?"


Sadly only 2.70€. No rings of power this time. Must keep trying...

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The Rabbit
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Will no one attempt the 12 bomb problem?
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scifibum
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I'm thinking about it, but having trouble so far. [Smile]
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The Pixiest
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A riddle?

http://tinyurl.com/6q9bct

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The Rabbit
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quote:
Originally posted by Tante Shvester:
Elmer's got it!

Good job. [Big Grin]


Edit:
Clarification --
The correct answer is hourglass.

I think it might be fair to claim that the cesium fountain (atomic clock) has more moving parts. Time is measured in the cesium fountain by monitoring the oscillation frequency of the electrons. I don't know how much Cs them is in an atomic clock, but even if there were only 1 nanogram, that would give you about 250 trillion electrons moving. To get that many 100 micron grains of sand you'd need over 250 cubic meters of sand.

And while all things have moving electrons, the cesium water clock is the only one I know of in which the movement of the electrons is the time measurement device.

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Flaming Toad on a Stick
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quote:
Originally posted by The Rabbit:
quote:
Originally posted by scholarette:
But from the first two weighings, you know if the bomb is too heavy or too light.

That's what I was looking for.


Now the harder puzzle. Same as above but you have 12 bombs.

I did mention that. I used the word "discrepancy".

12 bombs:1-12
M1: 1-4 vs 5-8.
If even, dud is in 9-12 group.
If so, 1-8 are regular. The second weighing will use one of the bombs 1-8 *say 1*.
M2: 1 and 9 vs 10 and 11.
If even, 12 is dud. You can use M3 to determine if it's lighter or heavier then any other bomb.
If uneven, M3 will be 10 vs 11. If 1+9 was heavier:
10=11, 9 is heavy dud.
10>11, 11 is light dud.
10<11, 10 is light dud.
If 10+11 was heavier,
10=11, 9 is light dud.
10>11, 10 is heavy dud.
10<11, 11 is heavy dud.
That's the easy part...

If M1 is uneven, dud is in 1-8. 9-12 are normal.
For M2, we use one of those bombs *say 9*. This part took me the greater part of an hour.

M2 1+2+5 vs 3+6+9.

From M1, 1-4>5-8, result A. 1-4<5-8, result B.
from M2, 1,2,5=3+6+9 Result C, 1,2,5>3+6+9 Result D, 1,2,5<3+6+9 Result E.
Cases:
AC, M3 is 7vs8.
7=8, 4 is heavy dud. 7>8, 8 is light dud. 7<8, 7 is light dud.
AD, M3 is 1vs2
1=2, 6 is light dud. 1>2, 1 is heavy dud. 1<2, 2 is heavy dud.
AE, M3 is 3vs12
3=12, 5 is light dud. 3>12, 3 is heavy dud.
BC, M3 is 7vs8
7=8, 4 is light dud. 7>8, 7 is heavy dud. 7<8, 8 is heavy dud.
BD, M3 is 1vs2
1=2, 6 is heavy dud. 1>2, 2 is light dud. 1<2, 1 is light dud.
BE, M3 is 3vs12
3=12, 5 is heavy dud. 3<12, 3 is light dud.

In case AE, 3<12 is not possible given the parameters of the question. In case BE, 3>12 is similarly not possible.

There goes an hour and a half. Thanks, Rabbit. [Mad]

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scifibum
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Nice work. I had the damnedest time figuring out how to do M2 in order to limit the possible results sufficiently.
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Flaming Toad on a Stick
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Like I said, the greater part of an hour. Gah!
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Nighthawk
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I got asked something similar to the "10 bomb" and "12 bomb" riddles in a recent job interview.

They used simple weights instead of high explosives, though.

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