Return of the Brain Eating Worms. AKA Ceeroque and Creyje, the Biological Trio. ALIAS

[Ray Jones]

Hmm. I'll post it.. later.

 

Hintus numeros unos:

 

 

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Psssst. Take 3 groups á 4 balls!

 

[DrMcCoy]

  Quote

Originally posted by Alien426

*picks up the M-16s from McCoy's very dead body*

 

*tries to say "THAT'S MY M-16, DAMNIT!"*

*but is dead, so nothing is coming out of his mouth*

 

...

[Ray Jones]

*kicks dead body, really hard and in the nuts*

[Alien426]

*keeps reading Stiff to get ideas of what to do with McCoy's cadaver*

[DrMcCoy]

*tries to whisper: "you could resurrect me..."*

*but is still dead*

[Ray Jones]

Maybe we should burn his corpse, huh?

[DrMcCoy]

*wants to scream: "NO!"*

*but is, yeah you guessed it, dead*

 

...

[Ray Jones]

Ok, you've heard what Thrik said. I need this thread to solve the riddle. And after that it will just fade away.

 

So BEHAVE.

 

Wait. I've already got the solution. OK. I will post it soon, this evening (CET), i think.

 

And after that it will just fade away.

 

Guess along, folks!!

[Skinkie]

I got it, you pick up allthe spheres one at a time and select the lightest one. This way you accomplish it with zero, I repeat, ZERO measurments. Give me a prize.

[Ray Jones]

OK, folks, here it is, my solution for the riddle and i found out on my own. And i cannot remember when i used the terms "equal" and "different balls" with so countless numberings this often the last time. ^_______^;;;;;;

 

I divided it into three spoilers so that, if maybe somewhere out there, someone, lonely and curious, someone like me, who just needs a hint to do the rest alone.. uh.. can do so. I mean, you should really try it. It's F.U.N.!

 

 

 

 

First weighting:

 

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Divide the balls into 3 groups A, B and C á 4 balls and weight groups A and B.

 

Possible results:

 

a) The balls of group A and B are equal: the sought-after ball is in goup C.

 

b) The balls of group A and B are different: the sought-after ball is in wether in group A or B.

 

 

 

Second weighting:

 

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case a) Take 3 balls of group C and replace them with 3 from group B on the balance.

 

Possible results:

 

a1) All balls on the balance are equal, again. The ball we're looking for is the left over one: now we already now which ball it is, but not wether if it's heavier or lighter.

 

a2) The balance shows a difference, so the sought-after ball is 1 of the 3 replaced balls: we don't know which ball exactly it is, but we know if it's heavier or lighter.

 

 

case b) Take 3 balls of group A from the balance, replace them with 3 balls from group B and fill group B up with 3 balls from group C (we know they're equal because the "right" ball is already on the balance).

 

Possible results:

 

b1) All balls are now equal. The ball we're looking for is 1 of the 3 from group A: as in result a2) we don't know which ball exactly it is, but we know if it's heavier or lighter.

 

b2) The weights are still different and the balance shows the same as in the first weighting: the 3 balls from group A we took out are equal to those from group B which replaced them and those again are equal the "new" ones from group C. The ball we're looking for must be 1 of the 2 from group A or B which haven't been exchanged. So we know it's 1 out of 2 and don't know if it would be heavier or lighter, so we have to keep in mind which one of the 2 balls is heavier (and which one lighter).

 

b3) The weights are still different, but have "changed sides": the ball we're looking for is 1 of the 3 from group B that have been moved from one side of the balance to the other. Again we don't know which ball exactly it is, but we know if it's heavier or lighter.

 

 

 

Third weighting:

 

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Case a2), b1), b3) Put 2 of the 3 balls on the balance.

 

Possible results:

 

I) Both balls are equal: the ball we're looking for is the third one. And according to the previous results we kow if it's heavier or lighter.

 

II) The balls are different: 1 of the balls on the balance is the sought-after and according to the previous results we kow if it's the heavier or lighter one.

 

 

Case a1) Take the already known to be different ball and weight it against an "equal one".

 

Result III): now we know if it's heavier or lighter.

 

 

Case b2) Take heavier one of the 2 balls and weight it against an "equal ball".

 

Possible results:

 

IV) The balls are equal: the left over ball is it. And it's lighter.

 

V) The balls weight different: it is the heavier ball. And it's err.. heavier.

 

 

 

You may now excuse me, I'll have a Martini or three to compensate my brain's rotation.

[Ray Jones]

Since i figured out that I like threads where I tend to talk to myself, here is yet another one. I expect the same participating like that i am used to.

 

Here it is.

 

There are 10 packs of coins á 10 coins. 9 of the packs consist of genuine coins, the coins of other one are bogus. We know the genuine coins weight 10g each and the false coins weight 11g each. Plus we've got a letter balacne and one try.

 

How to find the bogus pack-o-coins?

 

Go Ray. XD

[Alien426]

That's easy.

 

 

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You number the packs of cocains from 1 to 10 and take the corresponding number of coins from the pack.

 

1 coin from pack 1

2 coins from pack 2

3 coins from pack 3...

 

Then you weigh the whole heap and subtract 550 grams from the result. The difference is equal to the number of the pack with the counterfeit tickets for the Number Nine coins

 

[Alien426]

By the way, the previous puzzle wasn't solved. Hate to burst your bubble, but you started with "3 groups A, B and C á 4 balls" and the last weighing was with "2 of the 3 balls"...

[Ray Jones]

 

 

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When you have 3 balls and know that 1 of them is heavier (or in the other case lighter) and you put 2 on the balance, it's either the 1 of those on the balance or the other one. And you know if it's heavier or lighter.

 

 

[edit]

 

Ahh. You mean, FOUR and then THREE. Look at step 2.

[Alien426]

Try to crack this die hard puzzle:

 

You have a five gallon bottle and a three gallon bottle. You must put exactly four gallons of water in the bigger bottle.

[Ray Jones]

 

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Simon says I've seen the movie. But I already solved it before. So i'll make up a maths equation out of it.

 

x=3

y=5

 

4=4

4=2*2

4= 2(y - x)

y= x + 2

y= (2x + 4)/2

4= 2(((2x + 4)/2) - x)

4= 2((2x + 4)/2) - 2x

x= y - 2

4=2((2x + 4)/2) - 2(y - 2)

4=2x + 4 - (2y - 4)

4= 2x - 2y + 8

2= x - y + 4

-2= x - y

-2= 3 - 5 -> true

 

QED

 

 

 

 

 

/utter nonsense

[Alien426]

AFAIR it's never explained how they reach the solution in the movie.

[Ray Jones]

Hmm. Remember me, i can't, now that you mention this. Hmm-hmm. But they must have solved it, because in the end they marry.

[PoM]

You take the three galleon bottle and...

:c3po:

 

 

 

Aww!

Can't figure it out!

It's to hard!

 

Oh..wait..I know!

 

  Reveal hidden contents

x=3

y=5

 

4=4

4=2*2

4= 2(y - x)

y= x + 2

y= (2x + 4)/2

4= 2(((2x + 4)/2) - x)

4= 2((2x + 4)/2) - 2x

x= y - 2

4=2((2x + 4)/2) - 2(y - 2)

4=2x + 4 - (2y - 4)

4= 2x - 2y + 8

2= x - y + 4

-2= x - y

-2= 3 - 5 -> true

 

 

 

 

 

 

 

Oh yeah!

[Ray Jones]

Now that was.. err.. mmh.. ?

[Alien426]

Solution:

 

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Fill 3 gallon bottle (subsequently 3gb).

Put contents of 3gb in 5gb.

Fill 3gb.

Put contents of 3gb in 5gb. After 2 gallons it is full and there is 1 gallon left in the 3gb.

Empty 5gb.

Put contents of 3gb (1 gallon) in 5gb.

Fill 3gb.

Put contents of 3gb in 5gb.

 

Voilà: 4 gallons in the 5 gallon bottle.

 

 

[toenail1]

  Quote

Originally posted by Alien426

Try to crack this die hard puzzle:

 

You have a five gallon bottle and a three gallon bottle. You must put exactly four gallons of water in the bigger bottle.

 

My physics teacher gave us that as homework one night... he wasted a whole 10 seconds of my life that it took me to figure out (no, I haven't seen Die Hard either).