Brain Exercise...

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Sho
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Postby Sho » Mon Nov 07, 2005 4:29 am

My solution to the first puzzle (releasing it since there is now another puzzle to be worked on): 1, 3, 9, 27. The trick is that weights can be placed in the opposite pan to be used as "negative" weights. For example, place the 27 in the right pan and the 1 and 9 in the left pan to weigh out 17 units in the left pan.

In general, you can obtain the combinations for any system by adding 40 (1+3+9+27) to the weight wanted, writing this sum in base 3, subtracting 1 from each digit and taking the digits as the number of weights of the corresponding power of 3 wanted.

Example: You want to measure 17 units. 17+40=57, which is 2010 in ternary notation. This means you use (2-1)=1 weight of mass 27, (0-1)=-1 weight of mass 9, (1-1)=0 weights of mass 3 and (0-1)=-1 weight of mass 1.
In other words, the 27 on the right pan and the 9 and 1 on the left pan allows you to measure 17 on the left pan.


Seen AoM's puzzle before, but with a different setup.
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Pie
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Postby Pie » Mon Nov 07, 2005 4:45 am

well... how many beers did they have?
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Sho
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Postby Sho » Mon Nov 07, 2005 4:51 am

The number of beers they had is not needed.
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AoM
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Postby AoM » Mon Nov 07, 2005 5:00 am

but incidentally, it can be determined. 8)
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BlueNine
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Postby BlueNine » Mon Nov 07, 2005 4:13 pm

Ages are 9, 2 and 2.
They had 13 beers

OK a really easy one this time...

Emma, Laura and Abbie went parachuting. They jumped out of the airplane one after the other. This is what we know about their last jump :

Abbie landed before Emma
Emma jumped out of the airplane either before or after Laura and Abbie.
More girls jumped out of the plane before Abbie than landed before Emma.

In which order did the three girls jump out of the airplane and in which order did they land?
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SekoETC
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Postby SekoETC » Mon Nov 07, 2005 4:38 pm

Jumping: Emma Laura Abbie
Landing: Abbie Emma Laura

I'll post a new one in a moment.
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Sparkle
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Postby Sparkle » Mon Nov 07, 2005 7:07 pm

Wait a minute!! Both of you need to explain how you got your answers for those of us still sitting here looking at the screen like WTF!
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AoM
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Postby AoM » Mon Nov 07, 2005 7:43 pm

I'll let the others answer if they want to, but I'll give hints as to how to solve them.

In the riddle about the 3 children... write out the possibilities. There are only so many factors of 36. Find out what is possible and then figure out what the second and third clues are telling you.

In the parachuting riddle, draw out two 3x3 grids, one for landing and one for jumping. Each grid should have a label 1st, 2nd and 3rd on one axis, and Abbie, Emma, and Laura on the other. Looking at the clues, mark off the combinations that are impossible, and you can determine what the order is.
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SekoETC
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Postby SekoETC » Mon Nov 07, 2005 7:45 pm

More girls jumped out of plane before Abbie than landed before Emma


The chances are either "1 jumps before Abbie and 0 lands before Emma" or 2 jumps before Abbie and 1 lands before Emma".

(More than 2 can't be before anyone since there's just three participants.)

Abbie landed before Emma


Landing: Abbie Emma

This means at least 1 landed before Emma, which means it can't be 0, so it must be 1 which means 2 have jumped before Abbie.

Jumping: ? ? Abbie

Emma jumped out of the airplane either before or after Laura and Abbie


This means Emma can't be in the middle. Laura and Abbie are a bunch. Since we know that two people jumped before Abbie, the one who jumped before Abbie must be Laura and Emma jumped before both of them. Case closed.


But I don't get the red hat thing. Guys don't wanna wear red...
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Postby SekoETC » Mon Nov 07, 2005 8:25 pm

Ok I was trying to make my own riddle but it turned out too complicated, so I'll just throw a traditional one instead.

There are three-legged monster chickens hiding in the flock of normal innocent chickens. Now the farmer can count 77 legs and 34 heads. How many -? No, that would be too easy.
If the farmer shoots in the flock at random, what's the chance of hitting a mutant chicken instead of a normal one if we presume both kinds of chickens are of equal size, ain't dodging and the shot will certainly hit one and just one of the chickens?
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BlueNine
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Postby BlueNine » Mon Nov 07, 2005 8:39 pm

Ok first we start with the factorisations of 36: (36,1,1), (18,2,1), (9,4,1),
(9,2,2), (6,6,1), (6,3,2), (4,3,3).

Now the fact that there isn't enough info after Man #2 knows they're the same as the sum of the beers they've had means that the sum of the set of numbers can't be unique (or he'd have figured it out) so we see which ones have the same sum...

This leaves us with 9,2,2 and 6,6,1 (they both share the sum of 13)

Now the fact that there is an eldest means that it cannot be 6,6,1 or there would be 2 eldest...therefore it is 9,2,2

I hope I said that clear enough :D
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BlueNine
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Postby BlueNine » Mon Nov 07, 2005 10:19 pm

SekoETC wrote:Ok I was trying to make my own riddle but it turned out too complicated, so I'll just throw a traditional one instead.

There are three-legged monster chickens hiding in the flock of normal innocent chickens. Now the farmer can count 77 legs and 34 heads. How many -? No, that would be too easy.
If the farmer shoots in the flock at random, what's the chance of hitting a mutant chicken instead of a normal one if we presume both kinds of chickens are of equal size, ain't dodging and the shot will certainly hit one and just one of the chickens?


Well, 34 heads = 68 legs if all are normal. 77-68 gives us the amount of "extra" legs and as such the amount of monster chickens.

So out of the 34 heads, 9 of them are monsters...making the ratio of monster to normal 9:25.

This means that when the shot is fire there is a 36% (9/25 = 0.36) chance that the shot will hit a monster/mutant chicken.

Unless there's something I didn't get?
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AoM
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Postby AoM » Tue Nov 08, 2005 12:53 am

You got it. Give us a riddle.
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BlueNine
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Postby BlueNine » Tue Nov 08, 2005 1:18 am

Alrighty then...this one even has a bit of a Cantr feel about it...

Steven, Glen and Will work in a quarry on different shifts. The early morning shift, the late shift and the night shift. Each worker is given a different project. These are : extraction, preparation and sale of material. In no specific order, these three workers earn 1500, 1600 and 1700 dollars.

We also know the following:

The worker of the late shift earns less than Will.
Steven is assigned to the extraction of material.
The one who earns 1700 dollars, works a shift before the worker responsible for the preparation of material.
In the production sequence, the work done on the early shift comes before the work done by the worker who earns 1600 dollars.

How much does each worker earn ?
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AoM
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Postby AoM » Tue Nov 08, 2005 6:59 am

Glen works the late shift doing Preparation and earns $1500
Will works the night shift working in Sales and earns $1600
Steven works the early morning shift doing Extraction and earns $1700

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