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On December 06 2010 10:24 Trion wrote: A jug holds a quantity of water. Another jug holds an equal quantity of wine. A glass of water is taken from the first jug, poured into the wine, and the contents stirred. A glass of the mixture is then taken and poured into the water.
Is there more wine in the water or is there more water in the wine?
+ Show Spoiler + As long as the two jugs hold the same volume of liquid, then the amount of wine in the water jugmust be equal to the amount of water in the wine jug.
Logically, if there is X amount of wine in the water jug, then X amount of water must've been displaced into the wine jug.
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+ Show Spoiler + There is more water in the wine.
Treat the problem like this. say that the jug has 4 letters of fluid. so currently there is (A = water and B = wine)
AAAA = BBBB
when you take 1 glass of water say 2 letters and pour it into the wine you have
AA and BBBBAA
When you take some A/B mixture from that and put it in the water it is changed again.
Keep in mind that there are 1 parts water to 2 parts wine
AAB 1/2 A and BBBA 1/2 A There is more water in the wine. Edit, I'm stupid.
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On December 06 2010 10:54 Thoreezhea wrote:+ Show Spoiler + There is more water in the wine.
Treat the problem like this. say that the jug has 4 letters of fluid. so currently there is (A = water and B = wine)
AAAA = BBBB
when you take 1 glass of water say 2 letters and pour it into the wine you have
AA and BBBBAA
When you take some A/B mixture from that and put it in the water it is changed again.
Keep in mind that there are 1 parts water to 2 parts wine
AAB 1/2 A and BBBA 1/2 A There is more water in the wine.
Um ...
+ Show Spoiler + You say a glass of water = 2 letters? Then on the second pouring, you only scooped up 1.5 letters.
The mixture would be 2:1 B to A, so you'd be scooping up 4/3 letters of B and 2/3 letters of A, resulting in:
2 2/3A + 1 1/3B vs 2 2/3 B + 1 1/3 A.
Numerically, if you had 60 cc of water and 60 cc of wine, first scoop:
30 cc of water vs 60 wine + 30 water second scoop would contain 20 cc wine + 10 cc water resulting in:
40cc water, 20cc wine vs 40cc wine, 20cc water
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+ Show Spoiler [water and wine jugs] +The quantities of water in wine and wine in water are equal. This is due to the fact that there is half a glass of water and half a glass of wine (mixed) replaced for a full glass of water in the water jug.
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A farmer wants to cross a river with a fox, a sheep, and a cabbage. He must cross the river with a tiny boat. He could only take himself and one other - the fox, the goose, or the corn - at a time. He could not leave the fox alone with the goose or the goose alone with the corn. How does he get all safely over the stream?
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On December 06 2010 11:13 Trion wrote: A farmer wants to cross a river with a fox, a sheep, and a cabbage. He must cross the river with a tiny boat. He could only take himself and one other - the fox, the goose, or the corn - at a time. He could not leave the fox alone with the goose or the goose alone with the corn. How does he get all safely over the stream?
+ Show Spoiler +
Left bank: Fox, Corn | Boat: farmer, goose | right bank: 0 Left bank: Fox, Corn | Boat: farmer | right bank: goose Left bank: Fox | Boat: farmer, corn | right bank: goose Left bank: Fox | Boat: farmer, goose | right bank: corn Left bank: Goose | Boat: farmer, fox | right bank: corn Left bank: Goose | Boat: farmer | right bank: fox, corn Left bank: 0 | Boat: farmer, goose | right bank: fox, corn Left bank: 0 | Boat: 0 | right bank: fox, goose, corn
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Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural.
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On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural. Meh TL page is too narrow to write my proof.
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On December 05 2010 16:31 jeeneeus wrote:The blind man and checker board+ Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 + Show Spoiler + For part one, I think the answer is 8. There are 4 squares, each with a quarter that can be in one of 2 orientations. That means there are 2^4=16 different possibilities. Now, I had to create a methodical way to get every possible orientation. I find it's easiest if you think of the squares in a row, instead of in a square, so you can think of the orientations as a binary sequence. The original orientation would be written as 0000. Now you flip coins up in a ribbon sequence, meaning you go up to a certain point, then back down, and repeat until you get them all. So the first flip would be 0001. Then 0011 0010 0110 0100 0101 0111 1111 1011 1001 1000 1010 1110 1100 1101. This will cover every possible orientation (as proof, there are 16 of these, when you count 0000). Now you want to find the latest possible one with no earlier inverse (meaning there are 1's where there were 0's and vice versa). For example, the inverse of 1011 is 0100. This is because if there was an earlier inverse, it would have had all four coins with the same side, just the opposite side (heads instead of tails or vice versa). So if the coins were HTHH, although the flip 1011 would make them all Tails, the earlier 0100 would have made them all Heads. The latest of these is 0111, which is 8 total flips. Part two seems trickier, and I don't know if I wanna spend more time on it.
+ Show Spoiler + The first part is simple, as there's no possibility of making it under 7 turns, but there are 7 +6+5+4+3+2+1= 28 paths to do it, depending on the order of your flips. Do consider that if the coins starts as 4 heads/tails, it'll still count as if you've done it. Hence the 7 turns. (To make it more clear, if the coins starts as head, tails, head, tails, and you choose to flip the tails in this setup(whilst not knowing they were tails): the task would be done in the first turn. However, if they started out as head head tails head; it'd set them to head tails tails tails, in which you flip the head for that setup and clear the task in the second turn. Doing this in order so every possibility is given would take 7 turns whatever path you choose to take, as no path would make 2 or more setups to get cleared in one turn.
The thing about the second part is that when he turns the table, the possibilities of your path might or might not change. So out of these 28 paths you have to find the one that gets it right every turn no matter what order the coins are in at the time. Which to me sounds impossible. Therefore I figure it's a trick question, the blind man can tell which coins are heads or tails with his fingertips, so he only needs one turn.
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platypuses0
United States44 Posts
On December 06 2010 10:13 Trion wrote: How can half of 12 be 7? + Show Spoiler +12 in Roman numerals is XII, if you cut that in half horizontally you get VII which is 7.
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If I recall correctly there is a solution to the blind man even with a rotating board, but I'm not 100% sure it's the exact same moves. You simply need to flip certain coins such that the orientation won't matter for your next step, as in you continuously see different scenarios.
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edit reposted : please remove
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On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural. + Show Spoiler +x=0 y=0 z=0 n>2 0^n + 0^n = 0^n 0+0=0
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On December 06 2010 13:53 ieatpasta wrote:Show nested quote +On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural. + Show Spoiler +x=0 y=0 z=0 n>2 0^n + 0^n = 0^n 0+0=0
It says prove there ISNT a solution. You proved that there is one.
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Riddle: Poor people have it. Rich people need it. It is greater than God and more evil than the Devil. If you eat it you will die. What is it?
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United States4053 Posts
On December 06 2010 14:17 ieatpasta wrote: Riddle: Poor people have it. Rich people need it. It is greater than God and more evil than the Devil. If you eat it you will die. What is it? + Show Spoiler +
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On December 06 2010 14:17 ieatpasta wrote: Riddle: Poor people have it. Rich people need it. It is greater than God and more evil than the Devil. If you eat it you will die. What is it?
+ Show Spoiler + The answer is "nothing"
poor poeple have nothing rich poeple need nothing . nothing is greater than god . nothing is more evil than the devil . if you eat nothing you will die
Edit: needed to add spoiler tag
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How do you take 1 away from 19 and get left with 20 . Two possible answers(difficulty easy)
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On December 06 2010 14:40 MrProphylactic wrote: How do you take 1 away from 19 and get left with 20 . Two possible answers(difficulty easy) + Show Spoiler +XIX - I = XX Roman Numerals
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On December 05 2010 13:30 Dr. ROCKZO wrote:Here's one of my favourites. Logic: + Show Spoiler +
+ Show Spoiler +I already know the solution, but I was thinking about it, and...
...pragmatically, you have a 100% chance of NOT DYING (death being the number one "motivator" as far as riddles are concerned). You weigh a given bag, and if it's normal weight, eat the candy--if not, eat from any of the other bags haha.
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