Jump to content
Sign in to follow this  
Jayon

Puzzles

Recommended Posts

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:

 

trianglepuzzle.gif

 

  • Upvote 1

Share this post


Link to post
Share on other sites

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:

 

42puzzle.jpg

 

Share this post


Link to post
Share on other sites

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 biggrin.gif

Share this post


Link to post
Share on other sites

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? :)

Share this post


Link to post
Share on other sites

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

teapots.gif

 

By the way, I love tea smile.gif

Share this post


Link to post
Share on other sites

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

Share this post


Link to post
Share on other sites

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?

Share this post


Link to post
Share on other sites

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.

Share this post


Link to post
Share on other sites

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 tongue.gif

 

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.

Share this post


Link to post
Share on other sites

Okay, triple post (I'd be banned for this in a conventional forum biggrin.gif)

 

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?

Share this post


Link to post
Share on other sites

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 tongue.gif

 

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.

Share this post


Link to post
Share on other sites

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?

Share this post


Link to post
Share on other sites

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? tongue.gif

Share this post


Link to post
Share on other sites

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 smile.gif

Share this post


Link to post
Share on other sites

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

Share this post


Link to post
Share on other sites

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:

 

42puzzle.jpg

 

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.

Share this post


Link to post
Share on other sites
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?

Share this post


Link to post
Share on other sites

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 biggrin.gif

 

Two glasses

Share this post


Link to post
Share on other sites

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).

Share this post


Link to post
Share on other sites
(<-Canadian spelling).

 

You mean British spelling <.<

Share this post


Link to post
Share on other sites
(<-Canadian spelling).

 

You mean British spelling <.<

 

Probably; I just know that it's the way we spell it here, and it's not the way they spell it in the US. It gets kind of annoying when all of the word programs tell you you're spelling the word wrong -_- But anyway, back to the games...

Share this post


Link to post
Share on other sites
(<-Canadian spelling).

 

You mean British spelling <.<

 

Probably; I just know that it's the way we spell it here, and it's not the way they spell it in the US. It gets kind of annoying when all of the word programs tell you you're spelling the word wrong -_- But anyway, back to the games...

 

I don't get annoyed, if it counts for anything biggrin.gif I love that spelling.

Share this post


Link to post
Share on other sites

There's an "English (Canadian)" option too.

Share this post


Link to post
Share on other sites

So...

 

sudo.png

 

Sudoku Puzzle.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×