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#1 Jayon

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Posted 06 January 2010 - 10:40 PM

This topic is for all kinds of puzzles. There are no specific rules for what the puzzle should be, but I do have one requirement: the puzzle should have an actual logical answer. In other words, this is not for those riddles that are just word games. I also request that if you have the solution to a puzzle, that you give that answer as a spoiler. That way, anyone else who wants to figure out the puzzle can work on it without having the answer given away.

Here's my first puzzle. It takes a set of four shapes and puts them together to form a right triangle. The four pieces are then re-arranged, forming another right triangle with the same length and height, but the new triangle has a hole in it. How is this possible? Here is the graphic:

Posted Image


#2 Jayon

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Posted 13 January 2010 - 10:45 PM

Hmm, okay, not much in the way of a response. Perhaps this one will generate a little more interest; this is the 42 puzzle, designed by Douglas Adams for the Hitchiker's Guide to the Galaxy series. The answer, of course, is 42. The object of the puzzle is to find the question (there are several possibilities). Here it is:

Posted Image


#3 Chozo

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Posted 24 March 2010 - 03:12 AM

Douglas Adams is a bit beyond me, but the first I can definitely solve...the two triangular pieces don't have the same slope -- the smaller is 2/5 and the larger is 3/8 -- so the quadrilaterals formed by 1) the width of the smaller triangle and the length of the larger and 2) the width of the larger triangle and the length of the smaller have different areas, so the two Tetris rejects only fit inside one of them.

There's probably a more eloquent solution with the Fibonacci series, but...that's also a bit beyond me....
Wonderful puzzle, by the way Posted Image

#4 Jayon

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Posted 25 March 2010 - 01:27 AM

Woohoo! Someone finally responded to this topic! :P And you're right, the two triangles have slightly different slopes, which causes the difference. I'll leave the second puzzle un-answered, at least for now, as it has several possibly "solutions."

Got any puzzles of your own to throw out? :)

#5 Chozo

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Posted 25 March 2010 - 01:35 AM

If Teapot A can hold 10 cups of tea, then how much can Teapot B hold?
Posted Image

By the way, I love tea Posted Image

#6 snʞı

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Posted 25 March 2010 - 10:43 PM

I'd say it can only hold about one, considering how the pouring hole is so low on the side of the pot.

#7 snʞı

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Posted 25 March 2010 - 10:46 PM

I have one myself. Ok, here it goes:

You have 9 balls. All have the same weight except one which is heavier. You have a weighing scale, and in 3 turns you have to be able to determine which one weighs more. How do you do it?

#8 Chozo

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Posted 26 March 2010 - 01:53 PM

I'd say it can only hold about one, considering how the pouring hole is so low on the side of the pot.


Yep, you got it.

#9 Chozo

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Posted 26 March 2010 - 01:57 PM

I have one myself. Ok, here it goes:

You have 9 balls. All have the same weight except one which is heavier. You have a weighing scale, and in 3 turns you have to be able to determine which one weighs more. How do you do it?


Okay, here we go Posted Image

First turn: Put 4 balls on each side. If they are equal, then the unused ball is the heaviest. If not, discard the unused ball and the 4 balls on the lighter side.
Second turn: Put 2 balls from the heavy 4 on one side and 2 on the other. Discard the 2 balls on the lighter side.
Third turn: Put 1 ball from the heavy 2 on one side and 1 on the other. The heavier ball is the heaviest.

#10 Chozo

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Posted 26 March 2010 - 02:08 PM

Okay, triple post (I'd be banned for this in a conventional forum Posted Image)

You need to measure out 4 liters of [name of liquid here] but you only have 5-liter and 3-liter measuring cups. How do you do it?

#11 snʞı

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Posted 26 March 2010 - 09:42 PM


I have one myself. Ok, here it goes:

You have 9 balls. All have the same weight except one which is heavier. You have a weighing scale, and in 3 turns you have to be able to determine which one weighs more. How do you do it?


Okay, here we go Posted Image

First turn: Put 4 balls on each side. If they are equal, then the unused ball is the heaviest. If not, discard the unused ball and the 4 balls on the lighter side.
Second turn: Put 2 balls from the heavy 4 on one side and 2 on the other. Discard the 2 balls on the lighter side.
Third turn: Put 1 ball from the heavy 2 on one side and 1 on the other. The heavier ball is the heaviest.


Great! Now try it in two, which is the real puzzle.

#12 SEELE

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Posted 26 March 2010 - 11:31 PM

How about you put three balls in each side of the scale, with three extra leftover. If both groups being weighed are equal, the the leftover group contains the heavy ball. Weigh two balls from that group on both sides of the scale, with one leftover ball (again). If both balls are equal, then the leftover is the heavy ball.

Took like a minute of thinking so its probably wrong... (or I broke some rules).



You need to measure out 4 liters of [name of liquid here] but you only have 5-liter and 3-liter measuring cups. How do you do it?




I remember something like this a long time ago but it used different numbers I think. You fill the 5 liter cup completely and pour it into the 3 liter cup until the 3 liter cup is full. Then you throw out the liquid in the 3 liter cup, so you now have only two liters in the 5 liter cup. Pour the remaining 2 liters into the 3 liter cup, then refill the 5 liter cup completely. Then pour (carefully) the water from the 5 liter cup into the 3 liter cup untill its completely full. Since the 3 liter cup already had 2 liters in it, it needed only 1 more to be full. So that's 1 liter subtracted from the completely full 5 liter cup which leaves 4 liters.




Here's a lame one my math teacher told me.


You are a bartender. One day, an infinite number of mathematicians walk into your bar (this bar is in another dimension and can fit all of the mathematicians). The first mathematician asks you for one glass of beer. The second one asks you for half a glass of beer. The third one asks you for a third of a glass of beer. The fourth one asks for a fifth of a glass of beer. The sixth one asks you for a sixth of a glass of beer, and so on ad infinitum. How many glasses of beer must you pour to satisfy every single mathematician?

#13 Chozo

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Posted 27 March 2010 - 02:50 AM

You are a bartender. One day, an infinite number of mathematicians walk into your bar (this bar is in another dimension and can fit all of the mathematicians). The first mathematician asks you for one glass of beer. The second one asks you for half a glass of beer. The third one asks you for a third of a glass of beer. The fourth one asks for a fifth of a glass of beer. The sixth one asks you for a sixth of a glass of beer, and so on ad infinitum. How many glasses of beer must you pour to satisfy every single mathematician?


Well that's a harmonic series, so you would still have to pour an infinite number of glasses...it's got to be a trick question. Was "the fourth one asks for a fifth of a glass of beer" a typo, or is it part of the trick? Posted Image

#14 Chozo

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Posted 27 March 2010 - 03:00 AM

How about you put three balls in each side of the scale, with three extra leftover. If both groups being weighed are equal, the the leftover group contains the heavy ball. Weigh two balls from that group on both sides of the scale, with one leftover ball (again). If both balls are equal, then the leftover is the heavy ball.


Yeah, that's more or less right.

You put three on each side with three left over. If they are equal, then discard all six and weigh one and one (with one left over) from the remaining three -- if those are equal, the remaining one is the heaviest; otherwise, the one that was weighed heavier is the heaviest. If the original three and three are unequal, then discard the lighter three and the unused three and weigh one and one (with one left over) from the remaining three -- if those are equal, the remaining one is the heaviest; otherwise, the one that was weighed heavier is the heaviest.

I can't believe I didn't see that...good job Posted Image

#15 Chris

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Posted 27 March 2010 - 04:55 AM

1 to the first guy, cuz if I ran the bar I wouldn't be selling less than a whole glass of beer. >_>

#16 snʞı

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Posted 28 March 2010 - 01:11 AM

Hmm, okay, not much in the way of a response. Perhaps this one will generate a little more interest; this is the 42 puzzle, designed by Douglas Adams for the Hitchiker's Guide to the Galaxy series. The answer, of course, is 42. The object of the puzzle is to find the question (there are several possibilities). Here it is:

Posted Image


You know, now that I got a second look at it, perhaps the question could be:

How many randomly colored spheres can be placed in a 6 x 7 grid? The answer is 42.

#17 SEELE

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Posted 28 March 2010 - 02:45 AM

Well that's a harmonic series, so you would still have to pour an infinite number of glasses...it's got to be a trick question. Was "the fourth one asks for a fifth of a glass of beer" a typo, or is it part of the trick?




I must have shorted out my brain coming up with the answer to the last riddle. I wrote that down completely wrong... sorry.


HERE is the REAL puzzle/riddle/whatever:


You are a bartender. One day, an infinite number of mathematicians walk into your bar (this bar is in another dimension and can fit all of the mathematicians). The first mathematician asks you for one glass of beer. The second one asks for a half of a beer. The third asks for a fourth of a beer. The fourth one asks for an eight of a beer and so on and so on ad infinitum. How many glasses of beer must you pour to satisfy every single mathematician?

#18 Chozo

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Posted 28 March 2010 - 06:21 AM

HERE is the REAL puzzle/riddle/whatever:

You are a bartender. One day, an infinite number of mathematicians walk into your bar (this bar is in another dimension and can fit all of the mathematicians). The first mathematician asks you for one glass of beer. The second one asks for a half of a beer. The third asks for a fourth of a beer. The fourth one asks for an eight of a beer and so on and so on ad infinitum. How many glasses of beer must you pour to satisfy every single mathematician?


Okay, that makes things a bit easier Posted Image

Two glasses

#19 SEELE

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Posted 28 March 2010 - 07:31 PM

Well done. See I told ya it was lame. Most people in the class were blown away by it though.

#20 Jayon

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Posted 28 March 2010 - 09:51 PM

Yeah, reducing series like that (1+1/2+1/4+1/8+1/16...) tend to mess with your brain a little when you find out that the final infinite sum is actually a real number. It's pretty cool though :)

As for your response Ikus, that is indeed one of the possible answers :) There are others though. I'll give you guys a hint: think primary colours (<-Canadian spelling).