natsuki-chan-bg said: Ok so I should think about how can I prove that there are "n" blue eyed people, given the conditions here then? no, you're given that there are 'n' number of people what you have to do is figure out what happens once the mysterious visitor says that there is at least one blue-eyed person- remember, before this, the people on the island have no way of knowing if he or she has a blue eye |
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| One thing... you said they must leave immediately, but the only way I can work it out is if they have to leave after a set amount of time, let's say, after an hour passes. In which case the solution is 'n blue-eyed people leave after n periods of time'. |
| okay,,, i just wrote up my reasoning for this. Again, I'll assume that it's supposed to be a period of time, and not immediately (in which case the solution would only be 'n blue-eyed people leave', which I simply find strange) Anyway... Okay... to start with, let's assume there is two blue-eyed people (n= 2), B1 and B2. B1 doesn't know that he has blue eyes, but he sees one, and only one person with blue eyes: B2. B1 can't be sure that he has blue eyes with that knowledge though, so he won't leave the island (This also goes for B2). However, if B2 doesn't leave the island after hour 1, B1 knows that B2 sees someone else with blue eyes. Since B1 only sees one blue-eyed person, B2, he can reason that he must have blue eyes too. B2 can reason the exact same thing, and they both leave after hour 2, sicne they both found out that they are blue-eyed. 2 blue-eyed people leave after 2 days. Now, if there's 3 blue-eyed people, B3 can reason the same thing as in the above paragraph, and will expect both B1 and B2 to leave after hour 2. However, since neither B1 nor B2 leaves after hour 2, he understands that they see yet another person with blue eyes. That must, of course, be him, as he doesn't see any other blue-eyed people than B1 and B2. They have also reached the same conclusion, and they all leave after hour 3. 3 blue-eyed people leave after 3 hours. The same goes for 4, 5, 6... n blue-eyed people. |
Llama_Guy said: okay,,, i just wrote up my reasoning for this. Again, I'll assume that it's supposed to be a period of time, and not immediately (in which case the solution would only be 'n blue-eyed people leave', which I simply find strange) Anyway... Okay... to start with, let's assume there is two blue-eyed people (n= 2), B1 and B2. B1 doesn't know that he has blue eyes, but he sees one, and only one person with blue eyes: B2. B1 can't be sure that he has blue eyes with that knowledge though, so he won't leave the island (This also goes for B2). However, if B2 doesn't leave the island after hour 1, B1 knows that B2 sees someone else with blue eyes. Since B1 only sees one blue-eyed person, B2, he can reason that he must have blue eyes too. B2 can reason the exact same thing, and they both leave after hour 2, sicne they both found out that they are blue-eyed. 2 blue-eyed people leave after 2 days. Now, if there's 3 blue-eyed people, B3 can reason the same thing as in the above paragraph, and will expect both B1 and B2 to leave after hour 2. However, since neither B1 nor B2 leaves after hour 2, he understands that they see yet another person with blue eyes. That must, of course, be him, as he doesn't see any other blue-eyed people than B1 and B2. They have also reached the same conclusion, and they all leave after hour 3. 3 blue-eyed people leave after 3 hours. The same goes for 4, 5, 6... n blue-eyed people.  yea, i accidentally left out the time period part... my bad |
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| That was a important thing to leave out, actually =P Anyways... (I've used that word too much lately) One day, I'm out for a walk. On my way, I find something edible and even delicious. It has no bones, nor is it a fruit or vegetable. I decide to take it home I'm not hungry, but I want it for myself anyway, so I securely lock it up in a room in the basement, where I know for sure that no one can come in except me. However, when I come to get it a few days later, only remains are left. I don't walk in my sleep either. What did I find? |
| water? |
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| No. To be delicious it needs taste, something water doesn't have. |
| ice cream? (or popsicle) |
|
| Yes, ice cream perhaps? XP |
| I can't really be sure about the answer but I think there is ant/mice/pest that will eat your food so it could be a Sweet food or Cheese 8D |
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DeathkaiserG said: I can't really be sure about the answer but I think there is ant/mice/pest that will eat your food so it could be a Sweet food or Cheese 8D Agreed :D |
| //“The third time is for what will be. The way back will come but once. Be steadfast.” |
| @DeathkaiserG: Hmm... nice thinking ;) Anyone else have anymore ideas on what it is? :D |
| No, you don't really 'find' ice cream when on a walk, do you? And when I said nothing can get in, I did mean insects and animals (and disease) too, sorry for being unclear about that. |
| i find ice cream all the time when i walk around lol is it even a common food item? |
|
| Yes, but usually,you buy ice cream, you just don't find it somewhere =P And yes, it is a common food item. |
| i got it! it's an egg! the reason why you have only remains is because the egg hatched :D |
SaitoeOct 27, 2008 2:03 PM
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| thats right for sure :P only half an hour late to guess. |
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| Correct! Yay! *gives cookie* |
| awesome... okay, here's my next puzzle: You have a rectangular cake with a rectangular hole already taken out. Your task is to make one vertical cut that evenly divides the cake into two halves. This means that you can't simply cut horizontally through the cake. How do you determine how to make that single cut. Note: The hole can be any size, any place , any orientation within the cake. You may also measure any physical quantity of the cake and the hole. |
|
| I dont really understand the question, but I'll give it a go anyway. |
|
ASHGARY said: I dont really understand the question, but I'll give it a go anyway. naw, what i'm looking for is a general, mathematical statement that describes ANY situation. for example, how would you cut something like this precisely in half (ie, describe the cut line): |
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| If the geometric shape of the halves doesn't matter, then I presume the mass is the key. And you said that I can measure ANY physical quantity of the cake and the hole. So let's say that cake has n grams. Then I measure the hole, and it has x grams. So the remainder of the cake (when the rectagular hole taken out) has n-x grams. So if I can measure ANY quantity of the cake, I measure (n-x)/2 grams, and cut the cake on that line.... that way, I get two halves, but not symmetric in geometry, only halves in mass. Did I get it right? |
| //“The third time is for what will be. The way back will come but once. Be steadfast.” |
natsuki-chan-bg said: If the geometric shape of the halves doesn't matter, then I presume the mass is the key. And you said that I can measure ANY physical quantity of the cake and the hole. So let's say that cake has n grams. Then I measure the hole, and it has x grams. So the remainder of the cake (when the rectagular hole taken out) has n-x grams. So if I can measure ANY quantity of the cake, I measure (n-x)/2 grams, and cut the cake on that line.... that way, I get two halves, but not symmetric in geometry, only halves in mass. Did I get it right? you're getting warm. however, how do you know where that line is? remember, it has to be a straight cut |
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| What do I use to measure the cake? If it's a rectangular instrument, and I can move the cake left/right, I just have to move it until I reach the mass of (n-x) / 2. So I just use a knife then to cut it straight :D |
| //“The third time is for what will be. The way back will come but once. Be steadfast.” |
natsuki-chan-bg said: What do I use to measure the cake? If it's a rectangular instrument, and I can move the cake left/right, I just have to move it until I reach the mass of (n-x) / 2. So I just use a knife then to cut it straight :D hmm didn't think of it that way, but i guess it works... congrats what i was going for is you take the center of mass of the hole and the center of mass of the remaining piece and you draw a line connecting the two points. That is your cut line (it works, trust me) anyways, that's basically what your solution says! |
|
saitoXhenrietta said: natsuki-chan-bg said: What do I use to measure the cake? If it's a rectangular instrument, and I can move the cake left/right, I just have to move it until I reach the mass of (n-x) / 2. So I just use a knife then to cut it straight :D hmm didn't think of it that way, but i guess it works... congrats what i was going for is you take the center of mass of the hole and the center of mass of the remaining piece and you draw a line connecting the two points. That is your cut line (it works, trust me) anyways, that's basically what your solution says! Well I'd say that my solution works, but it's different. And yes, I believe that your solution works, but it's more mathematical than mine :D Ahhhh, I love this kind of riddles :DDD So I guess it's my turn then, but I'm gonna pass it to someone else who has a good riddle for us XD |
| //“The third time is for what will be. The way back will come but once. Be steadfast.” |
| Darn... natsuki-chan beat me to it =/ |
Llama_Guy said: Darn... natsuki-chan beat me to it =/ Sorry :PPP |
| //“The third time is for what will be. The way back will come but once. Be steadfast.” |
natsuki-chan-bg said: Not at all. So, what hard task have you got for us to solve?Llama_Guy said: Darn... natsuki-chan beat me to it =/ Sorry :PPP |
Llama_Guy said: natsuki-chan-bg said: Not at all. So, what hard task have you got for us to solve?Llama_Guy said: Darn... natsuki-chan beat me to it =/ Sorry :PPP I believe natsuki-chan deferred riddle-giving. anyone else have one? (i can't think of one right now...) |
|
| I'll do one then. It's child's play, really. A large freighter is anchored at a port. Its hull is black with small red stripes; a new stripe is positioned on the hull at every thirty centimeters of vertical height, and there are eight stripes in total, the first one barely touching the water. The tide is rising at a speed of 20 cm per hour. How long will it take until the water reaches the highest stripe? |
| never- the boat rises with da tide =D |
|
saitoXhenrietta said: Yeah... =Pnever- the boat rises with da tide =D |
| another relatively simple one: A large truck is crossing a bridge 50 miles long. The bridge can only hold 14000 lbs, which is the exact weight of the truck. The truck makes it half way across the bridge and stops. A small bird lands on the truck. Does the bridge collapse? Give a reason. |
SaitoeOct 28, 2008 1:07 PM
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| No, the truck is carrying the bird. The normal force exerted by the truck on the bird is canceling out the gravity pulling the bird to the center of the Earth. Since the bird is already very light in comparison with the truck and the weight of the bird decreases because of the normal power, the weight of the bird should be pretty much negligible. It's all a matter of physics. |
 |
Gurenn said: No, the truck is carrying the bird. The normal force exerted by the truck on the bird is canceling out the gravity pulling the bird to the center of the Earth. Since the bird is already very light in comparison with the truck and the weight of the bird decreases because of the normal power, the weight of the bird should be pretty much negligible. It's all a matter of physics. sorry, the normal force does balance out the gravity pulling the bird to the center of the earth. However, this means that there is an additional force on the truck downwards due to the bird's mass. Since the bird and truck are combined, they can be thought of as one newtonian object Think about it, if the laws of physics worked the way you described, then if you got on a scale, it would always weigh the same no matter how much extra weight you carry o.O I changed a couple of details to make the solution more reasonable, but it doesn't change the reasoning at all (small bird, 50 miles) |
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| The gas used by the truck after 25 miles weighs the same or more than the bird? |
Llama_Guy said: The gas used by the truck after 25 miles weighs the same or more than the bird? yep! |
|
| Here's the next one. Your task is to find out which numbers that go on the next row 2 12 1112 3112 132112 1113122112 311311222112 ? |
TheLlamaOct 28, 2008 2:40 PM
| 13211321322112 edit: I'll add an explanation to this. ^^. this doesnt have a thing to do with math, it's just logic. first row two. nothing to explain here. second row: one two. >> there is one two in the first row. therefore, the second row will have the numbers 12. third row: one one and one two. >> there is one one and there is one two. therefore, the third row will have the numbers 1112 fourth row: three ones and one two. >> there are three ones and one two. therefore, the fourth row will have the numbers 3112. etc... |
ASHGARYOct 28, 2008 4:12 PM
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| you beat me to it, ashgary XD |
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| mwahahaha >:D |
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| well, once we get confirmation (i'm 99.99999999999% sure you're right) it's your turn |
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| yeah I'm 1000% I'm right, but ima wait for llama to confirm anyway ^^. |
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| 1000% o.O |
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| typo-ness |
|
saitoXhenrietta said: Gurenn said: No, because the difference between your weight and the extra carriage is far smaller than the difference between the truck and the bird. Thus, you exert far less normal power than the truck.Think about it, if the laws of physics worked the way you described, then if you got on a scale, it would always weigh the same no matter how much extra weight you carry o.O |
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Gurenn said: No, because the difference between your weight and the extra carriage is far smaller than the difference between the truck and the bird. Thus, you exert far less normal power than the truck. that doesn't matter, physics just DOESN'T work that way... First of all, the according to the solution, the truck's weight before crossing the bridge is such that the addition of ANY weight, no matter how small in any form will cause the bridge to collapse. I was using a person carrying weight as an exaggerated extrapolation of your argument Secondly, normal forces simply cannot 'cancel' out weight. When the bird lands on the truck, it exerts a downwards force on the truck due to its weight. The truck exerts the normal force on the bird upwards. Notice that since the normal force acts on the bird, it does not magically make the truck any lighter. The truck itself exerts a net downward force on the bridge equal to its weight and the bird's weight, which is greater than the force the truck alone would have exerted. Please trust me on this, if not, then look it up or something XD |
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| You're wrong... Haha, just kidding =P Absolutely correct =D |
saitoXhenrietta said: http://en.wikipedia.org/wiki/Normal_forceGurenn said: No, because the difference between your weight and the extra carriage is far smaller than the difference between the truck and the bird. Thus, you exert far less normal power than the truck. that doesn't matter, physics just DOESN'T work that way... First of all, the according to the solution, the truck's weight before crossing the bridge is such that the addition of ANY weight, no matter how small in any form will cause the bridge to collapse. I was using a person carrying weight as an exaggerated extrapolation of your argument Secondly, normal forces simply cannot 'cancel' out weight. When the bird lands on the truck, it exerts a downwards force on the truck due to its weight. The truck exerts the normal force on the bird upwards. Notice that since the normal force acts on the bird, it does not magically make the truck any lighter. The truck itself exerts a net downward force on the bridge equal to its weight and the bird's weight, which is greater than the force the truck alone would have exerted. Please trust me on this, if not, then look it up or something XD Read the part on real-world applications. |
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