The worlds first riddle!

The bookworm 8"

The 'beer and whisky'; He drinks has first drink and puts the glass, upside own, over one of your glasses of whisky. You cannot drink that whisky with out touching his glass so he can take his time drinking the beer.

The straw I am not at all sure of - I'd need to experiment to see if this would work - but the way that I would try to do it would be to bend the straw such that the shortest 'leg' of the straw is just larger than the inside diameter of the bottle and then put he straw into the bottle 'bend first'

The residual spring in the straw will attempt to straighten it, but it cannot fully straighten as it is larger than the bottle, so - hopefully - it would jamb and the bottle would lift

Ahhh...yes it would.

At least, most of the time!

But I would suggest you try this several times when you are all alone so that you get the "bend" just right -- and so that you learn how to lift the bottle carefully enough so that you win your bet.

And don't be surprised if one of the prices of learning all this is an occasional lapfull of beer.

Oh, one other thing. There are straws -- and there are straws. Make damn sure you've got a straw that doesn't disintegrate at the drop of a hat.

(Swizzle sticks work also, but are very slippery!)

Not very often I will suggest this -- but on this one, you can take my word.

Lacomus wrote,
"your proof of '2 = 1':

At one point you divided by 'a - b',
but it was given that a = b
therefore a - b = 0, and dividing by zero is not allowed."

Go to the top of the class.

How about;
The following is what seems to be a mathematical proof that 10 equal 9.999999....
Is there anything wrong with it

a = 9.999999...
10a = 99.999999...
10a - a = 90
9a = 90
a = 10


"The bookworm 8"

The bookworm nibbles his way through approximately eight inches. The first page of the first volume is the rightmost page of the first volume. The last page of the last volume is the leftmost page of the last volume. Therefore, the only part of the first volume he eats is the front cover, and the only part of the last volume he eats is the back cover.

The reason the answer is not fourteen inches is that while each leaf is .01 inches thick, each leaf is two pages. So 400 pages is only two inches thick. Finally, I say approximately eight inches because it is a matter of semantics whether the bookworm ate the first page of the first volume.

The ?'Straw'

One way to lift the bottle is to bend the straw about two inches from one end. Insert the "V" shaped point formed into the mouth of the bottle until the straw can unfold. When you lift, the now unfolded portion will wedge in the body of the bottle below the neck. You can try this one for yourself easily- but practice with an empty bottle first. Some lightweight or paper straws may not have the rigidity necessary to keep from collapsing under the weight of the bottle plus six ounces of liquid.

The prisoner's defence.

This classic paradox tends to spark much debate when it appears. The judge clearly was correct: even if the prisoner had not been so confident in his defence, he would have been surprised by the execution. On the other hand, the defence seems to be logically well constructed. If on the last eve of the week the prisoner is still alive, there are not that many days left for the execution.

Let us simplify the paradox a bit- we do not need a whole week to cause a problem, two days will work just as well. If the judge tells the prisoner he is to be hanged either tomorrow or the next day, and he will not know until the day of the execution, we have a problem. After tomorrow, the prisoner will know the day, which violates the rule. So it must tomorrow. But if it must be tomorrow, then the prisoner knows the day. Which violates the rule? Must make the prisoner's head spin.

Which is fortunate for the judge, because it is this paradox he has constructed which makes the judge's sentence valid. He has created a logical system which traps the prisoner from being able to know the truth.

This even works if the judge had said: "You will be hanged tomorrow and you will not know that will be hanged tomorrow." Think about it- from the prisoner's perspective he cannot know that sentence to be true. If he could, it would be false.

Consider this parallel paradox (insert your name in the blank):
(Your Name Here) cannot correctly know this sentence to be true.
That sentence is true in all cases, yet whoever's name is in the blank cannot know it. Neat, huh?

I also agree with your reasoning in answer to Relative's questions, although, I don't think I would have come up with the answer myself.

BTW what was the answer to the number of cylinders question

Apart from one or two unanswered questions, that clears the decks for something a little different. (apart from still waiting for an answer to my riddle)

The riddle grows longer to make it easier.

We live not on land, or sea.
We have two eyes, but cannot see.
We close at night, and open at dawn.
Forget the Birds, and bring on the Bees
It makes reproduction a breeze
The closer you are, the more you reflect,
on whether, your answer can be correct.
Part above and part below
Wet and dry, don't you know.
But, at the end only lies.
We are


I was sent the seven problems below, and managed to answer two. Yes only 2! Ok, Ok, you can stop laughing now. Can you all do any better? I post it now, because I have at last been given the answers. Which I will of cause post in due time.

Here are seven tricky puzzletts, all dealing in some way with your friend, the number. Some are simple algebra problems, others involve numbers and logic, and a few wander completely out of the bounds of mathematics altogether into obscure trivia.

1. A length of a rope is 30 feet, plus half its total length. How long is the rope?

2. What's the next number in this series: 1,000; 1,000,000,000; 1,000,000,000,000,000,000,000,000,000; 100; 0; 4; 8; 3?

3. "Seven of Nine" is the curvaceous cyborg on Star Trek: Voyager. Who is "One of Six to Eight"?

4. Solve for x in this infinite summation: sqrt(x + sqrt(x + sqrt(x + ... ))) = 2

5. There are _ 1's _ 2's _ 3's _ 4's _ 5's _ 6's _ 7's _ 8's _ 9's and _ 0's in this sentence. Fill in the blanks with the correct digits to make the sentence true.

6. What's next: 1, 2, 5, 14, 42?

7. Quick! What is c in petasmoots per femtotribulation?

Try

The answer to the cylinders question is that you can have seven cylinders arranged so that every one of the seven is in direct physical contact with each of the others. I figured this out when I was looking for the rule that would prevent me from making the kind of dumb error that I made with the five coins. It requires very much the same kind of solution as the coins problem.

The '10 = 9.999...' proof:

I would take '10 = 9.999...' to be a true statement, so I didn't look any further.
Consider that 10 divided by 3 = 3.333...' and '3.333... times 3 = 9.999...' then 10 times 3/3 can equal either 10 or 9.999...
I think, however that I must have missed something as this seems too easy.

BTW, I don't seem to have received an answer to my question regarding the '10 coins in 3 glasses' problem (or if I have I was too dumb to see it).
I'll ask it again.
If the glasses are stackable, and if a glass holding a coin is stacked inside another glass, does the coin count as being in both glasses?

Try

Your seven questions:

I have an answer for 1, 4, and 5, and have a 'not entirely convincing' argument in support of my answer to 6.

Further information is required on 3, to whit, 'What is a Star Trek?'

Quick answer to 7: You do not make it clear whether 'c' refers to the speed of light, the abbreviation for 100, or what.
However. I reckon that the rest of it reduces to smoots over tribulations in Grouchos
On second thoughts, that could be re-written as Grauchosmoots over tribulation'.
The rest I will have another look at.

Hi,

Tryagain: 9.999* expansion is equal to 10 by definition. However, we can 'prove' it in the way you described. We can also think about it in terms of a limit of a series x(n+1)=x+9E-n .

About the mindreader : it was still an open issue, philosophically, at the time I read about this - and linked to one's belief in free will.
Since if you conclude that it's best to take only box B, and then change your mind to take also box A, how can your act of free will change the contents of box B? The same as prisoner's problem - you cannot trust the Alpha's claim if you beleive in free will.

Iacomus : you are correct about the beer. I left out one condition, though not absolutely necessary : the beer drinker gets to drink his first beer , then you start with whisky.

And I still can't see what is wrong with my solution to the straw problem? I'd put the beer together with a straw in a freezer. When the beer freezes, it is easy to lift the damn thing one foot! Even without straw touching the inside of the bottle altogether! When this solution occured to me, I was 100% this was correct

You have any idea of how low the temperature has to be in order to freeze a beer???


Where you gonna pull the freezer from?

Frank

Even without the assurance I would have entirely taken your word for the lifting of the bottle with the straw. Your post makes such good sense.

Frank: an open bottle of beer will freeze at -10 C. It is a CLOSED bottle that can't freeze, and this is because it is under a heavy pressure. Take a look here
http://www.darylscience.com/Demos/BeerFreeze.html

And around here it is very common to have a freezer in bars. They make ice with it.

Lacomus, Damn, No, I mean well done. That answer took me over a week!

I did post the above, above. However, I now see that it would be difficult to make it less clear. Your reply of 7. Are they all the same size

The full answer. odd number of coins?

Dividing the pennies among the glasses so each glass held an odd number of pennies is impossible. This can be demonstrated with the simple logic that any three odd numbers added together yield an odd number.

The trick, as you probably guessed, was to change the arrangement of the glasses so a penny could be in more than one glass. By placing an even number of pennies into a glass, an odd number of pennies into a second glass, and then placing the second glass into the first, both glasses contain an odd number of pennies.

One example, has two pennies in one glass, three pennies inside a glass within the first glass, and five pennies in the third glass.

Let me put the Beer saga to rest.

"I'll bet $25 I can lift that half full beer bottle one foot off the bar using only this plastic drinking straw. That's right- the only thing I'll touch the bottle with is this straw, and the only thing that will touch the straw other than the bottle will be my hand."

If anyone can get the bottle into the freezer only using the straw, then they win. Otherwise the answer as given is correct in theory and practice, and from what Frank said you will need a lot of practice. :wink:

a = 10
Actually there's nothing wrong with it. Ten does equal 9.99999..., as this proof clearly shows. Well spotted.

I can give no help with the seven questions. They are reproduced just as I received them. Good luck

Post up the answers as soon as you like. It will give others a chance to comment.

I rest my straw case.

Tryagain: I think I have 1,3,4,5,7.
Quite embarassed, but 2&6 still resisting!
And of course I'm not sure about 3&7..

Try

If you regard the cylinders as cigarettes all from the same packet I think you will then see it as I hoped I had explained it.:-)

Relative

While you might have difficulty in using your bottle-lifting technique, I agree that, IMHO, your answer fits all of the conditions of the question.

(Now let me tell you about boards over acid baths . . . Mwuhahahahaha!)

Relative

I think - I hope might be more truthful - that I did cover 'free will' and 'determinism' in that mindreader problem. But I do feel that the question as posed comes close to making predeterminism a given. The Centaurian is said to have carried this out many times and has not failed once. As it is either 'true' or 'false' each time then a long run of right guesses (or predictions) would take a lot of explaining away.

The 'beer and whisky' problem was easy enough even without the detail you did not include, once it was realised that no one can drink seven pints in the time it takes someone else to drink five shots. And the only other condition was 'do not touch the other's glass'.

Holy ****, Relative, they've got bars in Slovenia with ice machines that go down to -10 C!!!

Don't let our Mafia find out.

If they start importing them here, we're all in very bad trouble. :wink:

Just in case someone missed a classic:

There is a lamp in the attic that is controlled by one of the three on/off switches, located in the basement. From basement you cannot see the light when the lamp is on, and the only way to check it is to go upstairs and look.
Only one of the switches is correct, and they do have on/off labels on them.
Can you determine which switch controls the light, only going upstairs once?



And one for our beer saga:

Doing great business, the three of you - you and two partners - come across 21 barrels of beer;
Of which seven are full,
seven are half-full,
and seven are empty.

You decide to split the loot in three equal shares, consisting of equal amounts of beer AND barrels.
Without moving any beer out or into the barrels, how do you do it?

Frank, I must admit I don't have a clue, really, about bar-management. Neither do I have the skill you wrote about that is needed to lift a bottle with a v-shaped straw. I imagine one has to be pretty dextrous to do it, and I might persuade my friends to try it next time we have a drink.

Why would your mob be interested in ice-machines below -10C ? I might have some for sale

Uhh. I just won't stop..

You know, even with ordinary ice you can actually reach below -10, as a matter of fact.
You just add enough salt to the ice and voila!
Just be careful to wear gloves

Beer

If one takes a full barrel and the other two take two half full barrels each, there remains six full barrels and three half full barrels which divide easily between the three.

This gives all three the same amount of beer but two have one more barrel than the other. If they then let him take one empty barrel, this gives them the same amount of beer and the same amount of barrels each. They can each have two empty barrels from the six remaining.

So everyone now has the equivalent of three and a half barrels of beer and each has seven barrels.

Query about the switches:

Is the light 'on' when all three switches are on and 'off' when any one of them is off, or is it vice versa? Or does the state of one switch effect the state of the other switches? Or 'by majority'? Or any otrher option?

Jumping in too soon again! Given that I seem incapable of reading a question correctly first time, could someone/anyone tell me how I remove a post I have made?