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On April 19 2012 03:43 khanofmongols wrote:Hint: you can look at everyones eyes each day. Deduce in what situations people can leave given this. + Show Spoiler +The first day everyone knows that if there was only one person with green eyes then they would leave.
Since nobody leaves they know that there are at least 2 people with green eyes.
Now the second day happens and everyone knows that if there are only 2 people with green eyes then they will leave since they will see only one other person with green eyes and will know that they can leave. Since they don't leave, now there must be three people with green eyes.
This process continues until the 100 green eyed people leave on the 100th night.
Also this doesn't make sense because everyone knows how many others have green/blue eyes. Why would it take 100 days?
Something seems just not right with this answer srry...
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On April 19 2012 03:03 cmen15 wrote:Show nested quote +On April 19 2012 01:36 Geo.Rion wrote:damn, i arrived at the right conclusion, + Show Spoiler + 2, but didnt notice it has anything to do with circles, i simply noticed the numbers have a "weight" most of them 0, so i went on and discovered 6, 9, 0 had a "weight" of one, and 8 has "2". Wow... dude I like your way much better : )
I feel smart since I figured it out in 5-10 minutes but then I realised that I my mind is comparable to a 5 year old. Should I be proud or not... hmmm
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On April 18 2012 23:26 mentallyafk 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?
+ Show Spoiler +What will the other guard say if I ask him which door has the lion ?
Go through that door.
The idea is you are removing the uncertainity as you are sure there is one lie and one truth, so you negate the lie and get the answer.
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Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals.
The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live.
The chief agrees to give them one chance, but to live they have to solve a riddle.
He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that:
The man in front can't see any of the hats
The man in the middle can see the man in front's hat
And the man in the back can see the hats of the two others.
None of them can see the color of their own hat.
The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed.
Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
What color is his hat?
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I'm sorry if I posed the question unclearly. I'll edit in some extra information.
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On April 19 2012 03:52 Bubris wrote:
Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals.
The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live.
The chief agrees to give them one chance, but to live they have to solve a riddle.
He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that:
The man in front can't see any of the hats
The man in the middle can see the man in front's hat
And the man in the back can see the hats of the two others.
None of them can see the color of their own hat.
The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed.
Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
What color is his hat?
I actually had my geometry class reenact a similar problem to this while I was student teaching and I was demonstrating logic and reasoning. They totally loved it 
There was no cannibalism involved.
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On April 18 2012 22:47 yeaR wrote:
If Pinochio said my nose will grow right now, what would happen ?
+ Show Spoiler +I would look at pinochio's nose as a state based affect. When he says that his nose will now grow he is lying because nothing except him lying would have that affect. Because he is lying his nose would grow. At this point you would check to see if the parameters are still met 1- he lied and his nose grew and 2- he told the truth and his nose shrunk. If we assume a priority system then he would first increase in size and then decrease in size. Regardless I think this would only happen once. So there would be a growth and a decrease in size and then nothing would change.
On April 18 2012 23:02 kochanfe wrote:
You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
+ Show Spoiler +Fill the 5 gallon jug. Use the 5 gallon jug to fill the 3 gallon jug so you have 2 gallons left over. dump the 3 gallon jug. fill the 3 gallon jug w/ the 2 gallons from the 5 gallon jug. fill the 5 gallon jug.
Part 2- fill the 5 gallon jug. dump 3 gallons from the 5 gallon jug into the 3 gallon jug. dump the 3 gallon jug, fill the 3 gallon jug w/ 2 gallons from the 5 gallon jug. fill the 5 gallon jug. dump 1 gallon from the 5 gallon jug into the 3 gallon jug and you have 4 gallons left in the 5 gallon jug
On April 18 2012 23:26 mentallyafk 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?
+ Show Spoiler +"What would the other guard choose" and then go for the opposite. the liar would say that the truthful guard would always choose the lion door while the truthful guy would always say what the liar would say, ie choose the lion door. If you choose the opposite you will be rich
On April 19 2012 00:22 DarkPlasmaBall wrote:Riddle 3: An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man say to them? Solution: + Show Spoiler +Switch camels
+ Show Spoiler +I don't get why the camels would make a difference for each son. I would think that the wise man would have said that it doesn't matter how long it takes to finish but the sons may die otw.
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FREEAGLELAND26781 Posts
On April 19 2012 03:24 Bahamuth wrote:
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes:
- They are all perfect logicians
- Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour.
- This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain.
This was fun :3
+ Show Spoiler [solution] +Each green sees:
99g 100b
Each blue sees:
100g 99b
Day 1: There is at least one person that has green eyes.
Start from base case--
1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes.
Though not sure if correct.
Edit: hmm I feel slightly better after reading other posts in this thread about this.
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On April 19 2012 03:52 Bubris wrote:
Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals.
The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live.
The chief agrees to give them one chance, but to live they have to solve a riddle.
He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that:
The man in front can't see any of the hats
The man in the middle can see the man in front's hat
And the man in the back can see the hats of the two others.
None of them can see the color of their own hat.
The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed.
Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
What color is his hat?
+ Show Spoiler +
If the first two were both wearing hats of colour black, the third would have known that he is wearing red. So this possibility is ruled out. This means at least one of the first 2 is wearing a hat of colour red.
Deducing this, if the first person were wearing a hat of colour black, the second person would have known for sure that his hat is of colour red. So, this possibility is also ruled out.
The first person can deduce this, meaning, he is not wearing a hat of colour black. So the first person is wearing a hat of colour red.
Cant believe I solved it so soon :D
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On April 19 2012 03:52 Bubris wrote:
Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals.
The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live.
The chief agrees to give them one chance, but to live they have to solve a riddle.
He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that:
The man in front can't see any of the hats
The man in the middle can see the man in front's hat
And the man in the back can see the hats of the two others.
None of them can see the color of their own hat.
The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed.
Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
What color is his hat?
This really is a good one.
+ Show Spoiler +
If the 2 guys in the front both had a white hat, then the last guy would know he has a red hat. He doesn't know the colour of his hat, therefore, there is at least one red hat on the first 2 guys.
If the middle guy would see a white hat in front of him, he would know that he has a red hat (because there was at least one red hat between the two of them. He doesn't know the colour of his hat, therefore the first guy has to be wearing a red hat.
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On April 19 2012 03:58 flamewheel wrote:This was fun :3 + Show Spoiler [solution] +Each green sees:
99g 100b
Each blue sees:
100g 99b
Day 1: There is at least one person that has green eyes.
Start from base case--
1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes. Though not sure if correct. Edit: hmm I feel slightly better after reading other posts in this thread about this.
Yes sir, very well done! If you came up with it yourself, very well done.
One addition:
+ Show Spoiler +The people with blue eyes can just leave the day after the guys with green eyes do.
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On April 19 2012 03:52 Bubris wrote:
Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals.
The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live.
The chief agrees to give them one chance, but to live they have to solve a riddle.
He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that:
The man in front can't see any of the hats
The man in the middle can see the man in front's hat
And the man in the back can see the hats of the two others.
None of them can see the color of their own hat.
The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed.
Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
What color is his hat?
+ Show Spoiler +Red
The man in back doesn't see 2 black hats so says nothing.
The man in the middle doesn't see a black hat so he says nothing.
The man in the front can deduce that both men are seeing that he is wearing a red hat and can thus both also being wearing red hats so would be eaten if they said black.
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On April 19 2012 04:02 Bahamuth wrote:Show nested quote +On April 19 2012 03:52 Bubris wrote:
Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals.
The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live.
The chief agrees to give them one chance, but to live they have to solve a riddle.
He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that:
The man in front can't see any of the hats
The man in the middle can see the man in front's hat
And the man in the back can see the hats of the two others.
None of them can see the color of their own hat.
The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed.
Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
What color is his hat? This really is a good one. + Show Spoiler +
If the 2 guys in the front both had a white hat, then the last guy would know he has a red hat. He doesn't know the colour of his hat, therefore, there is at least one red hat on the first 2 guys.
If the middle guy would see a white hat in front of him, he would know that he has a red hat (because there was at least one red hat between the two of them. He doesn't know the colour of his hat, therefore the first guy has to be wearing a red hat.
Thats exactly what I said :D Though u seem to have confused black with white 
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On April 19 2012 04:04 Bahamuth wrote:Show nested quote +On April 19 2012 03:58 flamewheel wrote:This was fun :3 + Show Spoiler [solution] +Each green sees:
99g 100b
Each blue sees:
100g 99b
Day 1: There is at least one person that has green eyes.
Start from base case--
1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes. Though not sure if correct. Edit: hmm I feel slightly better after reading other posts in this thread about this. Yes sir, very well done! If you came up with it yourself, very well done. One addition: + Show Spoiler +The people with blue eyes can just leave the day after the guys with green eyes do.
How what if you have brown eyes? how does anything give you any information about that?
I.E. None of the blue eyes can be certain they have blue eyes just that they do not have green eyes.
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FREEAGLELAND26781 Posts
On April 19 2012 03:52 Bubris wrote:
Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals.
The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live.
The chief agrees to give them one chance, but to live they have to solve a riddle.
He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that:
The man in front can't see any of the hats
The man in the middle can see the man in front's hat
And the man in the back can see the hats of the two others.
None of them can see the color of their own hat.
The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed.
Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
What color is his hat?
+ Show Spoiler [solution] +Men are 123
Hats are B or R (2B 3R)
Combinations:
123
RBB
123
BRB
123
BBR
123
RRB
123
RBR
123
BRR
123
RRR
There are only two black hats. If we look at case 3 where the hat order is BBR if the guy in the back sees two black hats he knows he'll be wearing red. Thus everybody can be freed. Since in the problem the first person says he knows the answer (without anybody saying anything) this is not the case.
So this means that at least one of the front two people is wearing a red hat.
Since the third person hasn't said anything, the second person in line can deduce what I wrote above about somebody in the first two wearing a red hat.
Second person in line can logically deduce "if I see a black hat on person 1, then I am wearing red."
But since second person says nothing, this is not the case.
By thinking through what the second and third persons in line would think about, the first person in line can arrive at the conclusion that he is wearing a red hat.
Another fun one~
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FREEAGLELAND26781 Posts
On April 19 2012 04:04 Bahamuth wrote:Show nested quote +On April 19 2012 03:58 flamewheel wrote:This was fun :3 + Show Spoiler [solution] +Each green sees:
99g 100b
Each blue sees:
100g 99b
Day 1: There is at least one person that has green eyes.
Start from base case--
1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes. Though not sure if correct. Edit: hmm I feel slightly better after reading other posts in this thread about this. Yes sir, very well done! If you came up with it yourself, very well done. One addition: + Show Spoiler +The people with blue eyes can just leave the day after the guys with green eyes do.
Oh right >_>
Well there goes my high from getting a solution...
Edit: wait
On April 19 2012 04:06 whatwhatanut wrote:Show nested quote +On April 19 2012 04:04 Bahamuth wrote:On April 19 2012 03:58 flamewheel wrote:This was fun :3 + Show Spoiler [solution] +Each green sees:
99g 100b
Each blue sees:
100g 99b
Day 1: There is at least one person that has green eyes.
Start from base case--
1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes. Though not sure if correct. Edit: hmm I feel slightly better after reading other posts in this thread about this. Yes sir, very well done! If you came up with it yourself, very well done. One addition: + Show Spoiler +The people with blue eyes can just leave the day after the guys with green eyes do. How what if you have brown eyes? how does anything give you any information about that? I.E. None of the blue eyes can be certain they have blue eyes just that they do not have green eyes.
+ Show Spoiler +Oh right, this problem doesn't state that "there are only blue and green eyes", and that's not part of the information given to the inhabitants. So I guess in the end all the blue-eyed people would still be stuck.
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On April 19 2012 03:52 Bubris wrote:
Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals.
The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live.
The chief agrees to give them one chance, but to live they have to solve a riddle.
He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that:
The man in front can't see any of the hats
The man in the middle can see the man in front's hat
And the man in the back can see the hats of the two others.
None of them can see the color of their own hat.
The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed.
Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
What color is his hat?
+ Show Spoiler + So the back two do not know which hat they have. In order for the person in the back to know his color, the two in front of him would have to be black. Assuming the other two are logical people, they know that they don't have two black hats.
now that the man in the middle knows there is either 1 black and 1 red or 2 reds, in order for him not to know the person in front would have to be red.
Since the man in front can deduce these two occurances since none of them have spoken, he must be wearing a red hat.
I think that is it?
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Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
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On April 19 2012 04:04 Strykemard wrote:Thats exactly what I said :D Though u seem to have confused black with white
Ah you're right. I posted my solution without looking at yours obviously.
On the brown eyes, that's something I should've mentioned. I edited it in. It doesn't really change the idea of the riddle though.
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FREEAGLELAND26781 Posts
On April 19 2012 04:10 Go1den wrote:
Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
+ Show Spoiler [#1] +At first I thought in terms of mass distribution... Then I reversed the word not -_-
"ton"
+ Show Spoiler [#2] +The dog barks since he can hear the car coming? Or really if the driver is making a right turn already he wouldn't be crossing the middle of the intersection anyway.
+ Show Spoiler [#3] +Technically, the part defined as a hole has no dirt at all.
+ Show Spoiler [#4] +"new door" is an anagram for "one word".
+ Show Spoiler [#5] +Mathematically, as many times as you want since you can go negative. Yet literally, once you subtract 5 from 25 you can't do it again since you're subtracting 5 from 20, not subtracting 5 from 25.
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EPT![[image loading]](http://i.imgur.com/LYebe.png)
