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Darkcaster
United Kingdom11 Posts
A) the prince is 20 and the princess is 30 B)the prince is 30 and the princess is 20 C) the prince is 30 and the princess is 40 D) the prince is 40 and the princess is 30 E) they are both the same age. solutions + Show Spoiler + C) The prince is 30 and the princess is 40 30 + 40=70 , 70/2=35 , 35-10=25 , 25*2=50 , 50-10=40 princess=40 so prince=30 | ||
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ishboh
United States954 Posts
On May 13 2012 03:23 StoRm_res wrote: Maybe I'm just blind but I don't think this famous riddle has been posted yet Einstein's riddle 1. In a street there are five houses, painted five different colours. 2. In each house lives a person of different nationality 3. These five homeowners each drink a different kind of beverage, smoke different brand of cigar and keep a different pet. Who owns the FISH? HINTS 1. The Brit lives in a red house. 2. The Swede keeps dogs as pets. 3. The Dane drinks tea. 4. The Green house is next to, and on the left of the White house. 5. The owner of the Green house drinks coffee. 6. The person who smokes Pall Mall rears birds. 7. The owner of the Yellow house smokes Dunhill. 8. The man living in the centre house drinks milk. 9. The Norwegian lives in the first house. 10. The man who smokes Blends lives next to the one who keeps cats. 11. The man who keeps horses lives next to the man who smokes Dunhill. 12. The man who smokes Blue Master drinks beer. 13. The German smokes Prince. 14. The Norwegian lives next to the blue house. 15. The man who smokes Blends has a neighbour who drinks water. these are my favorite type of puzzles/riddles. you can logically come to a conclusion and it doesn't take wordplay, thinking outside the box, etc. solution: + Show Spoiler + 1-yellow, norwegian, dunhill, water, cats 2-blue, dane, blends, tea, horses 3-red, brit, pall mall, milk, birds 4-green, german, prince, coffee, FISH 5-white, swede, blue master, beer, dogs | ||
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ZapRoffo
United States5544 Posts
+ Show Spoiler + First we are going to use the initial hint, and say that we can express the submarine's position at any time t by saying it's: current position = initial position + time * speed. So there are two numbers, initial position and speed that we don't know but if we did, we could locate the submarine at any time we wanted. So the problem is, how can we account for (or count up) every possible initial position and speed without skipping any. Then we can just test them in that order until we eventually hit on the right one (which we will since speed and initial position are both finite once they are set). So we have to come up with a strategy to count every possible combination of two integers. That's what this image that Slithe posted is doing. Highlighting his strategy for hitting every combination of two integers (the x coordinate and the y coordinate of each of those points on the graph) that doesn't skip any, and you can continue following until you hit on the right two numbers for initial position and velocity. ![[image loading]](http://img167.imageshack.us/img167/24/infinityyd1.jpg) That's one system that works. The system I thought of before I saw his is to first count all the combinations that add up to 0--that is (0,0), then all the ones that add up to 1: (1,0) and (0,1) and then the corresponding combinations of those values with negative coordinates (-1,0) and (0,-1), then the ones that sum to 2: (2,0), (1,1), (0,2), and (-2,0), (-1,1), (-1,-1), (1,-1), (0,-2), then 3, etc... Given infinite time, this will hit on every combination of two integers and not skip any. So now we have an order of initial positions and velocities to test, we just have to say how we go about picking our bombing spot for each one. So we use the current position equation we started with: current position = initial position + time * speed. So our first test with my system is (0,0)--0 initial position and 0 speed, and we'll start at time 0. So it's current position if it started with those characteristics is 0 + 0*0 = 0, so we bomb position 0. Then we move on to (1, 0)--initial position 1 and speed 0, and now time has progressed one second so t = 1. Current position = 1 + 0*1 = 1. So we bomb position one to test for those starting characteristics. Next, (0,1) at t = 2 now, so current position = 0 + 1*2 = 2, bomb at 2. Next (-1,0) at t=3, so current position = -1 + 0*3 = -1, bomb at -1. Next (0,-1) at t = 4, so its current position = 0 + -1*4 = -4, so bomb at -4. And so on. Eventually we will land on the right combination and we will know exactly where it is because we know how much time has passed and we will hit it. And same thing for the general solution: + Show Spoiler + Now we ask, does this work if the submarine is in a 2-dimensional space, or a 3-dimensional space, etc.. The answer is yes. We just have to notice that we need more numbers to describe the initial position and velocity. For 2 dimensions we can say: Current horizontal (x) position = initial x + time * horizontal velocity and current vertical (y) position = initial y + time * vertical velocity. So now there are 4 integers which describe each possible starting state instead of two (initial x, initial y, x velocity, y velocity). And we can use the same counting strategy to count all of them without skipping. All combinations of adding up to 0--(0,0,0,0) which we test at time 0, (1,0,0,0) which we test at time 1, (0,1,0,0) at time 2, etc., systematically hitting the negatives for each coordinate as well as we go. And we can do the same thing for any number of dimensions. Does that make sense? | ||
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GoldenH
1115 Posts
You're saying that since there is a 1:1 mapping between X and f(X,t) that t doesn't matter. X = [x v] | f(X,t) = x+vt ______________________________ t=0 [0 0] | 0 t=1 [1 0] | 1 t=2 [0 1] | 2 t=3 [-1 0] | -1 t=4 [0 -1] | -4 t=5 [1 1] | 7 .... t=m [a b] | a + b*m and solving it for n dimensions would just be generalizing x to [x,y,z,...n] | ||
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Penke
Sweden346 Posts
On May 25 2012 01:34 Darkcaster wrote: A princess will be as old as a prince will be when the princess is twice as old as the prince was when the princess was half the sum of their current age. Which one of these statements is true? A) the prince is 20 and the princess is 30 B)the prince is 30 and the princess is 20 C) the prince is 30 and the princess is 40 D) the prince is 40 and the princess is 30 E) they are both the same age. solutions + Show Spoiler + C) The prince is 30 and the princess is 40 30 + 40=70 , 70/2=35 , 35-10=25 , 25*2=50 , 50-10=40 princess=40 so prince=30 Hmm, Isn't E) also a solution or am I thinking incorrectly? | ||
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Mindcrime
United States6899 Posts
On May 25 2012 01:34 Darkcaster wrote: A princess will be as old as a prince will be when the princess is twice as old as the prince was when the princess was half the sum of their current age. Which one of these statements is true? A) the prince is 20 and the princess is 30 B)the prince is 30 and the princess is 20 C) the prince is 30 and the princess is 40 D) the prince is 40 and the princess is 30 E) they are both the same age. solutions + Show Spoiler + C) The prince is 30 and the princess is 40 30 + 40=70 , 70/2=35 , 35-10=25 , 25*2=50 , 50-10=40 princess=40 so prince=30 This riddle appears in Baldur's Gate 2, so without even thinking about the math, I said to myself: + Show Spoiler + it's the third one  | ||
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ZapRoffo
United States5544 Posts
On May 25 2012 04:04 GoldenH wrote: Actually I think I understand it now. + Show Spoiler + You're saying that since there is a 1:1 mapping between X and f(X,t) that t doesn't matter. X = [x v] | f(X,t) = x+vt ______________________________ t=0 [0 0] | 0 t=1 [1 0] | 1 t=2 [0 1] | 2 t=3 [-1 0] | -1 t=4 [0 -1] | -4 t=5 [1 1] | 7 .... t=m [a b] | a + b*m and solving it for n dimensions would just be generalizing x to [x,y,z,...n] + Show Spoiler + It's not a 1:1 (bijective) mapping between X and f(X,t) (some positions can come up twice or not at all), it's between X and t (aka the natural numbers), but otherwise yes. | ||
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odeSSa
Sweden198 Posts
On May 25 2012 04:36 Mindcrime wrote: This riddle appears in Baldur's Gate 2, so without even thinking about the math, I said to myself: + Show Spoiler + it's the third one Exactly my thought  I learned it by trial and error in BG2 ^^ | ||
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nosliw
United States2716 Posts
You and your competitors are walking down the road with rocks and you can only carry one rock at a time. You cannot go backwards to pick a rock at the back. You exit the road whenever you want and the timer stops. You want to be the one with the biggest rock in the shortest amount of time. What is the statistical solution for the problem? I cannot find it on google for the life of me. Thanks, | ||
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ELA
Denmark4608 Posts
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South
80 Posts
On June 29 2012 11:12 ELA wrote: What concert would cost you 45 cents to attend? + Show Spoiler + 50 cent opening for Nickelback? lol | ||
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XxVenem94xX
United States41 Posts
Aw that's a good one I was going to say: + Show Spoiler + A 10% off 50 cent concert EDIT: Just noticed, it says Happy Birthday one my profile page, my birthday isn't for another hour and 20 minutes! Early Birthday FTW! ;D | ||
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NTTemplar
609 Posts
1/2 + Show Spoiler + On May 12 2012 00:02 AndyGB4 wrote: I read your last post before posting mine, and I disagree with your reasoning. You can't just say its 50% because they say "you reach into a random pocket"... The fact that you grabbed an Orange ball AFTER picking a random pocket affects the probability of which pocket you picked at random. Here are some perfectly good examples to prove my point: If the two pockets had 99 balls in each: P1 = 99 Orange balls P2 = 1 Orange ball, 98 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 99% chance you drew your ball from P1, There's a 1% chance you drew your ball from P2, therefore a 99%(99/100) chance the next ball will also be Orange. and a 1%(1/100) chance the next ball will be White. If the two pockets had 9 balls in each: P1 = 9 Orange balls P2 = 1 Orange ball, 8 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 90% chance you drew your ball from P1, There's a 10% chance you drew your ball from P2, therefore a 90%(9/10) chance the next ball will also be Orange. and a 10%(1/10) chance the next ball will be White. If the two pockets had 4 balls in each: P1 = 4 Orange balls P2 = 1 Orange ball, 3 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 80% chance you drew your ball from P1, There's a 20% chance you drew your ball from P2, therefore a 80%(4/5) chance the next ball will also be Orange. and a 20%(1/5) chance the next ball will be White. If the two pockets had 3 balls in each: P1 = 3 Orange balls P2 = 1 Orange ball, 2 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 75% chance you drew your ball from P1, There's a 25% chance you drew your ball from P2, therefore a 75%(3/4) chance the next ball will also be Orange. and a 25%(1/4) chance the next ball will be White. *** And this is the ACTUAL riddle: *** If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 66% chance you drew your ball from P1, There's a 33% chance you drew your ball from P2, therefore a 66%(2/3) chance the next ball will also be Orange. and a 33%(1/3) chance the next ball will be White. Under your logic from earlier, all these examples would still be 1/2 to you, correct? Don't you see how silly that logic looks on a larger scale though? Honestly, for anyone who still believes its 1/2, please try it out yourself, recreate this riddle with real objects. Take bottle caps, or pencils, or whatever, and split them up like in the problem, and record each draw. You will see that about 66% of the time, you will draw another Orange ball. The more times you do it, the clearer the probability will be. PLEASEEEEEE try it out  Explaination: + Show Spoiler + To take your first one with 99 balls in each pocket. You open P1, and now your chance for an orange ball is 100%, we seem to agree there, where you fall of is when we instead pick P2, and now the chance for an orange ball is also 100% After picking a pocket, if it contains an orange ball, we will pick it. Thus whetever it is 1 or a billion oranges in it is irrelevant as we are guaranteed to pick an orange one, since that is given in the riddle. Thus If the two pockets had 99 balls in each: P1 = 99 Orange balls P2 = 1 Orange ball, 98 White balls And you picket a RANDOM pocket and from it drew an orange ball, what is the chance for the remaining balls to be orange? There's a 50% chance you drew your ball from P1, There's a 50% chance you drew your ball from P2, therefore a 50% chance the remaining balls are Orange. and a 50% chance the remaining balls are White. Either you picked P1 or you picked P2, both will give you an orange ball. The chance of picking either P1 or P2 is 50%. In the original example we can exclude the double white one because a ball was drawn, if he had picked the double white pocket no balls would be drawn as there would be no orange ball to draw from it. | ||
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calderon
95 Posts
![[image loading]](http://msbrooks.com/blog/wp-content/uploads/2008/01/will_it_take_off.jpg) | ||
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Silentenigma
Turkey2037 Posts
On May 13 2012 03:23 StoRm_res wrote: Maybe I'm just blind but I don't think this famous riddle has been posted yet Einstein's riddle 1. In a street there are five houses, painted five different colours. 2. In each house lives a person of different nationality 3. These five homeowners each drink a different kind of beverage, smoke different brand of cigar and keep a different pet. Who owns the FISH? HINTS 1. The Brit lives in a red house. 2. The Swede keeps dogs as pets. 3. The Dane drinks tea. 4. The Green house is next to, and on the left of the White house. 5. The owner of the Green house drinks coffee. 6. The person who smokes Pall Mall rears birds. 7. The owner of the Yellow house smokes Dunhill. 8. The man living in the centre house drinks milk. 9. The Norwegian lives in the first house. 10. The man who smokes Blends lives next to the one who keeps cats. 11. The man who keeps horses lives next to the man who smokes Dunhill. 12. The man who smokes Blue Master drinks beer. 13. The German smokes Prince. 14. The Norwegian lives next to the blue house. 15. The man who smokes Blends has a neighbour who drinks water. it took me 10 minutes to solve  Solution: + Show Spoiler + After making a chart it shows german owns the fish | ||
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ELA
Denmark4608 Posts
Correct! | ||
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Zealot)KT(
Netherlands69 Posts
On June 29 2012 12:34 calderon wrote: if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off? + Show Spoiler + No. Seeing as the plane is constantly in one place, there is no friction with the air, and therefore there is no lift. | ||
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Grombidal
Sweden49 Posts
yes the plane will take off. Since the speed of the wheels have nothing to do with the rest of the plane, the plane will continue to travel forward. Mythbusters tested this in a episode. | ||
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Poltergeist-
Sweden336 Posts
On June 29 2012 12:34 calderon wrote: if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off? + Show Spoiler + At first I thought, no the plane will not take off. I assumed that it was the wheels propelling the plane forward but in reality the wheels just roll freely. This would mean that the plan would take off and the wheels woud just spin a lot faster than normal when the wheels are still touching the ground. Maybe it would take more power from the airplane to get it to lift than if the ground wasn't spinning underneath? I think..? | ||
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mikell
Australia352 Posts
why do you think planes accelerate instead of just taking off nonsensically? | ||
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