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Sanya12364 Posts
On April 19 2012 07:33 zJayy962 wrote: Ever play the game 24? Use the numbers 1, 5, 5, 5 and the four basic math operations ( + / - *) to get the answer 24. You can only use each number once. The operations can be used as many times as you want.
+ Show Spoiler +
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I'm not a regular on this forum so apologies if I'm quoting wrong :D
Solution discussion for Nehsb and Miicro: + Show Spoiler +I'm impressed :D Was that off the top of your heads or had you heard it before? It took me and my colleagues a while to work it out from a position of total ignorance. Nehsb is more correct though as he doesn't assume the starting state of the levers and he hints towards fully understanding the issue (but falls a little short). Basically there are two related facets:- You MUST flip a lever exactly once + There are 2 levers. What this effectively means is that the prisoners decide that the left lever is the "noop" lever. So the counter only cares about the right lever and you follow the logic you both described, so if a non-counter enters the room and the right lever is flipped to indicate someone needs counting then they throw away their flip on the left lever. You can think of this as changing the puzzle conditions to be having just one lever but flipping it is optional. Still, congrats if you worked this out from scratch!
Ninja Edit: Nehsb's solution accounts for the unknown starting state - good job! Miicro's solution accounts for the fact there are two levers - good job!
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On April 19 2012 00:08 DarkPlasmaBall wrote:Here's a riddle: How do you know when someone cheats and looks online for the solution to a riddle? + Show Spoiler +On April 18 2012 23:33 sc2system wrote:you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask? Took me like 2 minutes. Solution: + Show Spoiler + You ask "What would your brother say if I asked him behind which door the million dollars are?".
You will get the door where the dragon is.
Owned. I noticed that too.
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On April 19 2012 06:42 Diavlo wrote:I have a good one. You have twelve coins. One of them is slightly different in mass, either too light or too heavy. You have a two pan balance. How can you find out which coin is different and whether it's too light or too heavy in just three weighing? here is the answer: http://mathforum.org/library/drmath/view/55618.html
I don't like the answers they gave on that forum.
An easy way to do it would be to weigh coins one and two first. If they balance, then you know that neither of them is the coin you are looking for. In this case, remove coin 2 and replace it with coin 3 and repeat until you find a coin that doesn't balance with coin one. This is the coin you are looking for. The balance will tell you whether it is too light or too heavy.
If the initial weighing of coins 1 and 2 is not balanced, then you know that one of them is the coin you are looking for. You also know that the rest of the coins all weigh the same. Pick up coin two and set it aside. Put any other coin on the scale with coin 1. If it balances, you know that coin 2 is the coin you are looking for, and you can replace coin 1 with coin two to see if your coin is too light or too heavy. If it doesn't balance, then you know that coin 1 is your coin, and the scale will tell you if its too heavy or too light.
Those math forum guys made things way too complicated. (This is coming from an IDIOT) 
EDIT: I am a moron. I did not read the whole question. I wear women's underpants. My face is stupid. I am a poop-nosed butt sniffer. The end.
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What would a NSFW riddle look like?
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On April 19 2012 08:08 CrazyF1r3f0x wrote: What would a NSFW riddle look like? mind. blown
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On April 19 2012 08:08 CrazyF1r3f0x wrote: What would a NSFW riddle look like? roflmao
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On April 19 2012 08:05 MichaelDonovan wrote:Show nested quote +On April 19 2012 06:42 Diavlo wrote:I have a good one. You have twelve coins. One of them is slightly different in mass, either too light or too heavy. You have a two pan balance. How can you find out which coin is different and whether it's too light or too heavy in just three weighing? here is the answer: http://mathforum.org/library/drmath/view/55618.html I don't like the answers they gave on that forum. An easy way to do it would be to weigh coins one and two first. If they balance, then you know that neither of them is the coin you are looking for. In this case, remove coin 2 and replace it with coin 3 and repeat until you find a coin that doesn't balance with coin one. This is the coin you are looking for. The balance will tell you whether it is too light or too heavy. If the initial weighing of coins 1 and 2 is not balanced, then you know that one of them is the coin you are looking for. You also know that the rest of the coins all weigh the same. Pick up coin two and set it aside. Put any other coin on the scale with coin 1. If it balances, you know that coin 2 is the coin you are looking for, and you can replace coin 1 with coin two to see if your coin is too light or too heavy. If it doesn't balance, then you know that coin 1 is your coin, and the scale will tell you if its too heavy or too light. Those math forum guys made things way too complicated. (This is coming from a mathematician)  Did you read the part where you only have three weighing?
The answer posted on the math forum is the correct one.
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On April 19 2012 08:26 Diavlo wrote:Show nested quote +On April 19 2012 08:05 MichaelDonovan wrote:On April 19 2012 06:42 Diavlo wrote:I have a good one. You have twelve coins. One of them is slightly different in mass, either too light or too heavy. You have a two pan balance. How can you find out which coin is different and whether it's too light or too heavy in just three weighing? here is the answer: http://mathforum.org/library/drmath/view/55618.html I don't like the answers they gave on that forum. An easy way to do it would be to weigh coins one and two first. If they balance, then you know that neither of them is the coin you are looking for. In this case, remove coin 2 and replace it with coin 3 and repeat until you find a coin that doesn't balance with coin one. This is the coin you are looking for. The balance will tell you whether it is too light or too heavy. If the initial weighing of coins 1 and 2 is not balanced, then you know that one of them is the coin you are looking for. You also know that the rest of the coins all weigh the same. Pick up coin two and set it aside. Put any other coin on the scale with coin 1. If it balances, you know that coin 2 is the coin you are looking for, and you can replace coin 1 with coin two to see if your coin is too light or too heavy. If it doesn't balance, then you know that coin 1 is your coin, and the scale will tell you if its too heavy or too light. Those math forum guys made things way too complicated. (This is coming from a mathematician)  Did you read the part where you only have three weighing? The answer posted on the math forum is the correct one.
No. I skipped that part. My mistake. I feel like such an IDIOT right now. Excuse: Haven't slept in 30 hours. Excusable? You decide.
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On April 19 2012 07:54 keplersfolly wrote:I'm not a regular on this forum so apologies if I'm quoting wrong :D Solution discussion for Nehsb and Miicro: + Show Spoiler +I'm impressed :D Was that off the top of your heads or had you heard it before? It took me and my colleagues a while to work it out from a position of total ignorance. Nehsb is more correct though as he doesn't assume the starting state of the levers and he hints towards fully understanding the issue (but falls a little short). Basically there are two related facets:- You MUST flip a lever exactly once + There are 2 levers. What this effectively means is that the prisoners decide that the left lever is the "noop" lever. So the counter only cares about the right lever and you follow the logic you both described, so if a non-counter enters the room and the right lever is flipped to indicate someone needs counting then they throw away their flip on the left lever. You can think of this as changing the puzzle conditions to be having just one lever but flipping it is optional. Still, congrats if you worked this out from scratch! Ninja Edit: Nehsb's solution accounts for the unknown starting state - good job! Miicro's solution accounts for the fact there are two levers - good job!
I'd heard it before. There's also a really hard version that I'm curious if anyone has an easier solution to:
20 prisoners are in jail who are all gathered in the main hall and addressed by the warden who explains the following. "In 1 hour you will all be placed in solitary confinement and never see each other again. You will individually be escorted by the guards to the "Lever Room" in which the are a pair of levers that have no function except being able to be flipped either Up or Down. You can flip any number of levers, including all of them or none of them. We make no guarantees about the order you will be called to the Lever Room in, and we may take any of you there more than once. If any of you prisoners tell the guard "We have all been to the lever room" and this is true you will all be released, otherwise you will all be shot."You do not know the starting position of the levers. You are allowed to write down a strategy and give it to all of the prisoners, but the strategy each prisoner gets must be the same; you are not allowed to "pick a leader" or anything like that.
The changes are in bold: - Still 2 levers - You can now flip any number of levers - Every prisoner must have the same strategy - You do not know the starting states of the levers
The problem gets much easier if you do know the starting state of the levers.
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On April 18 2012 23:33 sc2system wrote:you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask? Took me like 2 minutes. Solution: + Show Spoiler + You ask "What would your brother say if I asked him behind which door the million dollars are?".
You will get the door where the dragon is.
Ask "Which one of you is lying?" ^that assumes i'm asking them both the question simultaneously, which i am, right?
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On April 19 2012 08:56 Pocketpurple wrote:Show nested quote +On April 18 2012 23:33 sc2system wrote:you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask? Took me like 2 minutes. Solution: + Show Spoiler + You ask "What would your brother say if I asked him behind which door the million dollars are?".
You will get the door where the dragon is.
Ask "Which one of you is lying?" ^that assumes i'm asking them both the question simultaneously, which i am, right? I assumed you can only ask one. Even if you could ask the same question to both, they will both say the other is lying. In any case, you still don't know which door is which.
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On April 19 2012 08:44 Nehsb wrote: [snip lever riddle]
The problem gets much easier if you do know the starting state of the levers. + Show Spoiler +The solution is actually exactly the same, you just have to count through everyone twice instead of just once. Perhaps you can also speed up the process by using the second lever somehow but I don't care to figure out if that's possible.
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On April 19 2012 08:56 Pocketpurple wrote:Show nested quote +On April 18 2012 23:33 sc2system wrote:you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask? Took me like 2 minutes. Solution: + Show Spoiler + You ask "What would your brother say if I asked him behind which door the million dollars are?".
You will get the door where the dragon is.
Ask "Which one of you is lying?" ^that assumes i'm asking them both the question simultaneously, which i am, right?
They'd just say the other one is lying.
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On April 19 2012 09:25 crate wrote:Show nested quote +On April 19 2012 08:44 Nehsb wrote: [snip lever riddle]
The problem gets much easier if you do know the starting state of the levers. + Show Spoiler +The solution is actually exactly the same, you just have to count through everyone twice instead of just once. Perhaps you can also speed up the process by using the second lever somehow but I don't care to figure out if that's possible.
Are you taking in account the everyone must have the same strategy restriction?
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On April 19 2012 09:34 Nehsb wrote:Show nested quote +On April 19 2012 09:25 crate wrote:On April 19 2012 08:44 Nehsb wrote: [snip lever riddle]
The problem gets much easier if you do know the starting state of the levers. + Show Spoiler +The solution is actually exactly the same, you just have to count through everyone twice instead of just once. Perhaps you can also speed up the process by using the second lever somehow but I don't care to figure out if that's possible. Are you taking in account the everyone must have the same strategy restriction? nevermind didn't see that part
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United States7483 Posts
On April 19 2012 07:37 TanGeng wrote:Show nested quote +On April 19 2012 07:33 zJayy962 wrote: Ever play the game 24? Use the numbers 1, 5, 5, 5 and the four basic math operations ( + / - *) to get the answer 24. You can only use each number once. The operations can be used as many times as you want. + Show Spoiler +
+ Show Spoiler + Gah ignore me, my dislexia strikes again!
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What English word retains the same pronunciation, even after you take away four of its five letters?
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United States7483 Posts
On April 19 2012 09:45 ExcisionExecute wrote: What English word retains the same pronunciation, even after you take away four of its five letters?
+ Show Spoiler +
Here's one:
What word is the only word in the English language to have 3 double letters in a row? Example: Balloon has 2 double letters in a row.
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