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Avon
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Right, here I go.

Before the performance Beethro and Bombus number the 24 permutations of the numbers 1,2,3,4
0: (1,2,3,4)
1: (1,2,4,3)
2: (1,3,2,4)
...
23: (4,3,2,1)

After the audience members have chosen their cards, Beethro considers the four blocks of numbers, 1-24, 25-48, 49-72, 73-96. The audience members have chosen three cards, so Beethro can choose a block which does not contain any of the audience members' cards.

Suppose Beethro chooses a card from this block. Then if he imagines the four chosen cards arranged ascending order, and imagines he writes '1' on the the first auidence member's card, '2' on the second, '3' on the third and '4' on his own he will have a permutation of the numbers 1,2,3,4. Note that since none of the audience members chose a card from Beethro's block, this permutation does not depend on which card Beethro chooses from his block.

This permutation has a corresponding number in the range 0-23. Since Beethro's block consists of 24 consecutive numbers, Beethro can choose a card from his block so that the sum of the four chosen cards modulo 24 is the number of this permutation.

When Bombus gets the four cards he arranges them in ascending order and finds their sum modulo 24. He then works out the corresponding permutation and hence determines the order in which the cards were chosen.

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06-13-2004 at 01:29 PM
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agaricus5
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What about cards larger than 96?

Does this method count for those as well?

I'm not going to make a suggestion myself as I haven't got a suitable puzzle.

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06-13-2004 at 02:23 PM
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Avon's answer is correct. Well done!

Agaricus: The method works for any set of three cards. If one of the cards have a number greater than 96, more than one of the blocks will contain no cards selected by the audience, but this doesn't cause any trouble for the method, as it just means that Beethro have more than one way to pick his card.

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06-13-2004 at 03:00 PM
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Avon
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Here's my puzzle.

Eliza has a secret place where she stores her collection of sweets, bottle caps and pennies. However this place isn't as secret as Eliza thinks because her brother Lester knows where it is.

On the first of January, Eliza had 33 sweets, 48 bottle caps and 59 pennies in her collection. She went to check on it a few times during January, but didn't add any new items. Several times each day Lester went to Eliza's secret place and took two items of different types and added an item of the third type, so that her stash doesn't shrink too quickly. For example, on one visit he might have taken a bottle cap and a penny and added a sweet.

On the first of February, Eliza was shocked to find that she had only one type of item in her stash. What type was it?


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06-14-2004 at 02:07 AM
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The_Red_Hawk
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Pennies?

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06-14-2004 at 02:22 AM
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Avon
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The_Red_Hawk wrote:
Pennies?
Yes, pennies.

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06-14-2004 at 02:29 AM
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Two immortal beings, Pierre and Jacques, were playing a game. There were ten squares in a row on the board. For one turn, the person drew five cards from a standard deck. If three or four cards were either from the same suit or with the same value, a "bink" was scored, for three points. If five of them filled those conditions, a "jink" was scored for five. There are no other methods of scoring. The game is only over if someone scores three jinks in a row, and the person with the highest score wins. When three or five points are scored, the person moves his counter (for Pierre, a monkey head, for Jacques, a soup can) that many squares. Jacques decided to spook out his opponent, and started boasting that he had played an infinite number of games before. He also said that there was no point value that he had not reached before. Pierre, however, was sure that this was impossible. He knew that of course there is no end to numbers, but it was always possible that Jacques had gotten huge numbers of points before. He simply had to prove that he could find all the numbers that Jacques couldn't have gotten and he could prove himself to be the math whiz that he said himself to be. He tried to work it out on paper, picking ridiculously large random numbers, but they were all possibgle. Every day after that, he spied out Jacques' house and looked at the records of his games. It seemed he had had all numbers before, but they might have been forged. Pierre worked late into the millenium trying to figure it out. He even tried to use electrodes on Jacques' cranium to try and read his thoughts. But he couldn't do anything.

Can you help Pierre?


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06-14-2004 at 02:55 AM
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eytanz
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Well, there are plenty of small numbers Jacques couldn't have reached - say, 2 or 23 - but I'm not entirely clear if that answers the puzzle or not.

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06-14-2004 at 03:12 AM
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DiMono
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Any number that doesn't satisfy 3n + 5m would, by definition, be unreachable. I think it's easier to say it's not possible to have a negative score, or 1 or 2 points.

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06-14-2004 at 03:47 AM
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Avon
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Avon wrote:
The_Red_Hawk wrote:
Pennies?
Yes, pennies.
:w00t Sorry, I wasn't saying pennies was the answer. I thought your question meant 'Do you really mean pennies? What kind of girl collects pennies? And what kind of brother steals pennies from his sister? I mean, they're pennies!' My answer was meant to mean 'Yes, I do mean pennies. My sister's the kind of girl who would collect pennies, and I'm the kind of brother who would steal them from her. I mean, they're pennies!' Also, I would have expected some reason why 'pennies' was the answer.

I guess I'm to blame for the misunderstanding, and the rules don't cover this situation.

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06-14-2004 at 03:54 AM
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DiMono
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Oh. In that case, it would be bottle caps.

For one item to be left, there would have to be 1 of two piles left to take from at once. Since all the piles just switch between even and odd numbers every time, the only way for this to happen is for pennis and sweets to be at 1 each, and bottle caps to be at some even number.

Since I don't have a puzzle ready, let's just go with Hawk's puzzle if I'm right.

[Edited by DiMono on 06-14-2004 at 03:13 AM GMT]

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06-14-2004 at 04:11 AM
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eytanz
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DiMono wrote:
Any number that doesn't satisfy 3n + 5m would, by definition, be unreachable. I think it's easier to say it's not possible to have a negative score, or 1 or 2 points.

Ah - I was wrong about 23, of course; I misread the instructions as saying you quit after the third jink, not the third jink in a row.

I'm guessing that the puzzle is to list all positive integers that cannot be reached by Jacques. The answer is, quite simply - all the numbers that are too small for the game to end yet. Since the game requires at least 15 points to quit, scores of 14 or less are impossible. For scores over 15, we must find all the scores that can be made with a combination of at least 3 successive jinks.

A score of 16 is unreachable, as is a score of 17, as to end the game with 3 jinks you'd have to start with eithe 1 or 2 points which is impossible. 18 is possible - get a bink, then 3 jinks. 19 is again impossible. 20 is impossible because you have to get 4 jinks in a row to get it, but the game would end after the third. 21 is possible (2 binks, 3 jinks). 22 is impossible (you need 7 to start).

23 is possible - get a jink, a bink, and 3 jinks.
24 is possible - 3 binks, 3 jinks

25 is impossible (you need 5 jinks in a row).

26 is possible - a jink, 2 binks, 3 jinks.
27 is possible - 4 binks, 3 jinks.
28 is possible - 2 jinks, a bink, 3 jinks.

Now you have a sequence of three possible scores in a row, so all scores greater are possible - just add binks accordingly.

Therefore, the highest unreachable score is 25. Jacques is clearly a liar. What would it matter since the game is based entirely on luck and not on skill is a different question.

---

Assuming I didn't miss anything, I'll just resubmit Avon's puzzle as my own to solve that mess (I know the answer to it, so I can judge it if he doesn't want to).


[Edited by eytanz on 06-14-2004 at 09:46 AM GMT]

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06-14-2004 at 04:32 AM
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eytanz
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Oh dang, we've got a bit of a gift of the magi situation here between me and DiMono :)

Not sure what to do now - DiMono, do you want to post a puzzle?

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06-14-2004 at 04:34 AM
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DiMono
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Like I said in my post, I don't have one ready, so I left it to Hawk's puzzle, which you solved, so you're up! Thanks for waiting the couple minutes for me to post though, which allowed continuity :) Cheers

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06-14-2004 at 06:33 AM
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The_Red_Hawk
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25 is impossible (you need 5 jinks in a row).


You can always get 5 jinks that aren't in a row. You could have 2 jinks, a no-scorer, 2 more, a no-scorer, and 1 more.



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06-14-2004 at 05:19 PM
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eytanz
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Oh - I misunderstood, then - I thought that "a row" meant only among the scoring turns. Anyway, 20 and 25 are possible, then, so that largest impossible number is 22.

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06-14-2004 at 05:31 PM
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Pinnacle
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WARNING: This is a sadistic puzzle.

You're standing next to a door. On the other side of the door is an infinitely large room. To your right is an infinite number of balls, labeled 1,2,3,etc. At 11:52, you throw balls 1 and 2 into the room. Instantly, ball 1 comes back. At 11:56, you throw 3 and 4; 2 comes back. This goes on repeatedly at 11:58, 11:59, 11:59:30, 11:59:45, etc. the time half of the previous. At 12:00, how many balls are in the room?

There are 5 possible answers, but they're all flawed.




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06-16-2004 at 06:10 PM
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agaricus5
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Well, assuming that you can move infinitely fast, by 12:00, you will have a semi-paradox. There will be an infinite number of balls in the room since you will need an infinite number of halves of the remaining time to get there, and so an infinite number of throws, an infinite number of balls still in the pile, and an infinite number of returned balls. The strange thing is that the pile will still have an infinity of balls which is an infinity smaller than it started, and the number of balls that have returned, although an infinite number itself, should be half the number in the room, which is also an infinite number.

The puzzle breaks down at infinity, unfortunately, so I don't know if there really is an answer.

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06-16-2004 at 06:20 PM
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Yes. 'Tis a bad puzzle IMHO. Best that I can say is the for the purposes of the balls, 12.00 will never be reached.

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06-16-2004 at 07:31 PM
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DiMono
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masonjason wrote:
Yes. 'Tis a bad puzzle IMHO. Best that I can say is the for the purposes of the balls, 12.00 will never be reached.
Gah, you got here before me. This is actually a well-known paradox, the inventor of which I cannot recall the name of. The common version is that you can never reach your destination, because first you have to go half way, and then before you can traverse the remaining distance you have to go half of it, and so on...

----------------
xxxxxxxx--------
xxxxxxxxxxxx----
xxxxxxxxxxxxxx--
xxxxxxxxxxxxxxx-
and so forth toward very small distances, but since you're only ever travelling half the remaining distance, you never actually reach your destination, just become infinitely close.

I'll take the less expected answer and say that the first ball is never thrown in. Let's suppose you arrive at the door at 11:50. Before 11:52 can roll around, you need to reach 11:51. But before 11:51 can happen you need to see 11:50:30, and before that you must see 11:50:15. Continuing, we see that the first ball never gets thrown, because time actually doesn't move.

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06-17-2004 at 05:20 AM
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agaricus5
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That is also not really true, I think.

It really depends on your time-frame. What you say only works if you are watching time slow down in a way where an amount of time relative to you becomes half of what it previously was to the clock every time the time-period finishes. You could argue that one will never see 11:51 in real time (that's 1 percieved second/the time it takes a light beam to travel 299792458 metres in a vacuum) but this can happen only if you look at it in halving intervals of time, which obviously requires a changing shift in the observer's measurement of the length of time.

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06-17-2004 at 02:23 PM
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I'm going to have to say there are 42 balls in the room.


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06-17-2004 at 04:51 PM
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Pinnacle wrote:

WARNING: This is a sadistic puzzle.

You're standing next to a door. On the other side of the door is an infinitely large room. To your right is an infinite number of balls, labeled 1,2,3,etc. At 11:52, you throw balls 1 and 2 into the room. Instantly, ball 1 comes back. At 11:56, you throw 3 and 4; 2 comes back. This goes on repeatedly at 11:58, 11:59, 11:59:30, 11:59:45, etc. the time half of the previous. At 12:00, how many balls are in the room?

There are 5 possible answers, but they're all flawed.



I'm just gonna spit another flawed answer out, saying that because you can't actually reach 12:00 there will actually be 0 balls in the room, because the balls return instantly eventually, so as you approach the limit, any finite number of balls will have returned before noon.
Answers: Infinity, Infinity/2, 0, 42, and.... my final flawed answer is 1 because of the final pair you threw at noon, which you never would, 1 instantly returned, while the other is still there... all previous balls returned, somehow, despite the fact that it seems you are constantly increasing the number of balls.


Let's stick to concrete mathematics from here on out, eh?

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06-17-2004 at 05:06 PM
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Hmmm... you have an infinite amount of balls in that starting room... That does need infinite space, doesn't it? So you have two infinite-space-rooms... But the space is of no use since you spend the rest of your life (That's actually 8 minutes but seems like eternity) throwing balls like a maniac...

Man, that does even make Cantor dizzy!

Pinnacle, will you provide us with that sadistic solution or another puzzle? Of should I come up with one?


06-27-2004 at 10:09 PM
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RonnnL wrote:

Pinnacle, will you provide us with that sadistic solution or another puzzle? Of should I come up with one?


Isn't infinity fun?
MUWAHAHAHAHAHAHA!!!!! :lol

Fine.
Click here to view the secret text




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06-28-2004 at 03:45 PM
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So, um, who gets to step up to the mound next?

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06-28-2004 at 04:58 PM
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There will be a finite number of balls in the room, because even if you've got infinite strength and speed, some time before 12:00 you will reach the critical speed that generates enough heat to evaporate both you and your arm (unless it gets ripped off you by some freak vortex first). But I'm no physics expert. ;)

- Gerry
07-08-2004 at 01:18 AM
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Okay, almost a full month and no one has submitted a puzzle. I guess I'll get it started again.

You are in a room that is an 8x8x8 perfect cube. There are no windows, or doors (don't ask me how you got in there!) In the center of the floor there is a 12 inch pipe that is sticking 6 inches out of the floor. In the bottom of the pipe is a ping pong ball with a diameter that is one millimeter smaller than the inner diameter of the pipe. The only items in the room with you are a 12 inch piece of string, a match, a magnifying glass, a 6" ruler and a paper clip. How do you get the ping pong ball out of the hole?

Please refer to the rules on the first page of this thread if you have forgotten them. Especially, if you know the answer then please wait a day for others to try to puzzle it out...

Game on,

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08-05-2004 at 08:13 PM
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Schik
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This is somewhat disgusting, but if you urinate or spit enough into the pipe, the ping pong ball will raise with the... fluid... and you can grab it.

Better wash your hands afterwards, though.


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08-05-2004 at 10:29 PM
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If that's not what you're looking for, you can just blow into, or across the top of, the pipe. Moving air exerts less pressure than still air, so the ball should rise if you blow hard enough.

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08-05-2004 at 10:35 PM
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