The riddle thread - Page 3
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Pawsom
United States928 Posts
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LTT
Shakuras1095 Posts
On March 15 2009 08:21 fanatacist wrote: Five men walk up the mountain. The person in front is six years older than the person in the back. The person walking second is half of three times the front person's age. The person walking in third is the product of the age of the first and last person, divided by the half of the age of the second person, rounded to the nearest digit. The age of the fourth person is the difference between the last person's age and the third person's age, multiplied by 3. The sum of the last person's age and the third person's age is 79. If all the men are aged between 30 and 50, what is the age of each one? EDIT: Sorry for the multiple edits, had to fix it because I miscalculated something. There is something wrong with your math. Label the people A,B,C,D and E in order. Sentence 1 means A = E + 6 Sentence 2 means B = 3A/2 Your range is 30-50 The Minimum age is 30, which means the lowest possible age for A is 36. That means the lowest possible age for B is 54 which is outside your range. | ||
Archaic
United States4024 Posts
On March 15 2009 08:05 Hot_Bid wrote: ![]() Lmao, I love you Hot_Bid | ||
paper
13196 Posts
On March 15 2009 11:07 LTT wrote: There is something wrong with your math. Label the people A,B,C,D and E in order. Sentence 1 means A = E + 6 Sentence 2 means B = 3A/2 Your range is 30-50 The Minimum age is 30, which means the lowest possible age for A is 36. That means the lowest possible age for B is 54 which is outside your range. yeeeeeeeeah haha i was doing this earlier and i was like : | | ||
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Ares[Effort]
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DEMACIA6550 Posts
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yB.TeH
Germany414 Posts
1coin from the 1st machine, 2coins from the 2nd, 3from 3rd, and so on.. just take all the coins you got, 210??? and calculate how much it should weigh, if 1ounce less its the 1st machine, if 2 the 2nd and so on | ||
thunk
United States6233 Posts
On March 15 2009 08:45 imBLIND wrote: A man lives on the 30th floor of an apartment building. Everyday when he returns home from work, he takes the elevator. If he is with another person, then he goes straight to the 30th floor and to his apartment building. If he is by himself, he goes to the 28th floor and walks up the last two flights of stairs to his apartment. Why? imDUMB. What's the answer to this one? | ||
Ozarugold
2716 Posts
+ Show Spoiler + The guy is a midget and can only reach the buttons for the 28th floor. If someone else is there, he asks them to press the 30th floor button for him. | ||
imBLIND
United States2626 Posts
Or they're really good at using google lol. my riddles have been answered correctly like 4 different times by different ppl =p + Show Spoiler + On March 15 2009 10:17 Klive5ive wrote: + Show Spoiler + Does he take a bag and place it face down on the floor. Then he rubs the stone in the bag against the floor (assuming he's standing on a stone stage like an amphitheater). The stone will crumble on the outsides creating a layer of white rubble. Then he simply removes the stone from the bag and it appears to be white. This one is wrong though =p | ||
Llamaz
Australia90 Posts
On March 15 2009 05:56 EffOrt wrote: 1.In this thread you will write some riddles and in spoiler write the answer 2.If you have a riddle but do not know the answer also post that in bold or make it stand out some way 3.If you're going to answer someone else's riddle please put it in spoiler btw this is just a example so don't take it to seriously ^^ example 1.What can you catch but you can't throw? + Show Spoiler + a flu example 2.What came first the chicken or the egg? Don't know #1. You are wearing a blindfold and thick gloves. An infinite number of quarters are laid out before you on a table of infinite area. Someone tells you that 20 of these quarters are tails and the rest are heads. He says that if you can split the quarters into 2 piles where the number of tails quarters is the same in both piles, then you win all of the quarters. You are allowed to move the quarters and to flip them over, but you can never tell what state a quarter is currently in (the blindfold prevents you from seeing, and the gloves prevent you from feeling which side is heads or tails). How do you partition the quarters so that you can win them all? http://en.wikipedia.org/wiki/Chicken-and-egg_problem Definitions The dilemma can be interpreted differently using different definitions of a chicken or an egg. In biology, the term egg is biologically ambiguous and the theory of punctuated equilibrium, for example, does not support a clear division between a chicken and the closest ancestors of that chicken. Both of those factors tend to contribute to the circular nature of the question (causing problems similar to either a hasty generalization or a fallacy of definition). Below are a few different definitions that could be assumed and their logical outcomes.[6] If the egg is not necessarily of any specific type: Then it could be said that the egg came first, because other animals had been laying eggs long before chickens existed, such as the dinosaurs. In biology, egg is used as a general term in this way. *If only an egg that will hatch into a chicken can be considered a chicken egg: Then a re-consideration of the original question suggests: Some animal other than a chicken laid the first chicken egg which contained the first chicken. In this case the chicken egg came before the chicken. In reality, many scientific theories suggest that this would not have been a simple event. For example, the theory of punctuated equilibrium theorizes that the actual speciation of an organism from its ancestral species is usually the result of many mutations combined with new geographical surroundings, called cladogenesis. If only an egg laid by a chicken can be considered a chicken egg: Then a re-consideration of the original question suggests: The first chicken (which hatched from a non-chicken egg) laid the first chicken egg. In this case the chicken came before the chicken egg. Again, this would not necessarily be a straightforward event. | ||
TimeShifter
Singapore235 Posts
On March 15 2009 06:31 Klive5ive wrote: + Show Spoiler + Djabenete is right though, technically if you grabbed 20 coins they are all heads anyway. You could grab any number of coins and as long as you flipped 20 over you would win. The infinity thing doesn't really add anything to the riddle. + Show Spoiler + yes it does, because 20/infinity = almost almost almost zero thats why you can pick up most coins and get heads in math u learn to assume anything/infinity = 0 | ||
imBLIND
United States2626 Posts
Afterwards, they also noticed that the gun was missing. The dogs could not smell it and the metal detectors could not find it. They know its a handgun or revolver and that it is in the house. All they know about the murderer is that he is a dancer. Where is the gun? | ||
TheosEx
United States894 Posts
On March 15 2009 14:27 Ozarugold wrote: + Show Spoiler + The guy is a midget and can only reach the buttons for the 28th floor. If someone else is there, he asks them to press the 30th floor button for him. + Show Spoiler + I was about to say that it was because he was banging some chick that he didn't want his wife to know about. | ||
fanatacist
10319 Posts
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deathgod6
United States5064 Posts
On March 15 2009 16:00 imBLIND wrote: A man was shot multiple times in a house. There were 10 holes in the victim's body; 4 were exit wounds. The bullets were nowhere to be seen in the room even though there were visible bullet holes and ricochets. Where did the bullets go? Afterwards, they also noticed that the gun was missing. The dogs could not smell it and the metal detectors could not find it. They know its a handgun or revolver and that it is in the house. All they know about the murder is that he is a dancer. Where is the gun? Wait.. is the victim the dancer or the murderer a dancer? | ||
imBLIND
United States2626 Posts
On March 15 2009 16:41 deathgod6 wrote: Wait.. is the victim the dancer or the murderer a dancer? sorry, murderer. Forgot an extra -er =p | ||
Scaramanga
Australia8090 Posts
On March 15 2009 13:16 EffOrt wrote: you have 20 coin machines, each of which produce the same kind of coin. you know how much a coin is supposed to weigh. one of the machines is defective, in that every coin it produces weighs 1 ounce less than it is supposed to. you also have an electronic weighing machine. how can you determine which of the 20 machines is defective with only one weighing? (by one use, we mean you put a bunch of stuff on the machine and read a number, and that's it -- you not allowed to accumulate weight onto the machine and watch the numbers ascend, because that's just like multiple weighings). you are allowed to crank out as many coins from each machine as you like. Easy You label the machines 1-20, from machine 1 you take 1 coin, from machine 2 you take 2 all the way up to twenty, you then place all these coins on the weighing machine and then however many ounce's it is out is the machine that is dodgy, so if its 3 ounces out its machine 3 that is bad | ||
15vs1
64 Posts
On March 15 2009 08:21 fanatacist wrote: Five men walk up the mountain. The person in front is six years older than the person in the back. The person walking second is half of three times the front person's age. The person walking in third is the product of the age of the first and last person, divided by the half of the age of the second person, rounded to the nearest digit. The age of the fourth person is the difference between the last person's age and the third person's age, multiplied by 3. The sum of the last person's age and the third person's age is 79. If all the men are aged between 30 and 50, what is the age of each one? EDIT: Sorry for the multiple edits, had to fix it because I miscalculated something. If i get everything right the only solution for age of the second man is 60 and it is out of range. So it seems there is something wrong with conditions. I have more interesting math question. We have an n-digit number aaaa.....aaa (it can be 2222, 5555555 etc). The question is how many digits a will appear in all numbers from 1 to our number aaaa.....aaa. Example: Lets say a = 1 and n=2. In this case our number is 11. All number from 1 to 11 are: 1 2 3 4 5 6 7 8 9 10 11. There are four digits 1 (1, 10 and 11). Of course answer should be expressed in terms of a and n. | ||
Loanshark
China3094 Posts
On March 15 2009 16:00 imBLIND wrote: A man was shot multiple times in a house. There were 10 holes in the victim's body; 4 were exit wounds. The bullets were nowhere to be seen in the room even though there were visible bullet holes and ricochets. Where did the bullets go? Afterwards, they also noticed that the gun was missing. The dogs could not smell it and the metal detectors could not find it. They know its a handgun or revolver and that it is in the house. All they know about the murderer is that he is a dancer. Where is the gun? + Show Spoiler + 4 exit wounds = 4 more entry wounds, for a total of 8 holes so far. The man's ass and mouth count for the other 2 holes. The bullets punctured the walls and flew out windows so they are not in the room. Am i rite | ||
fanatacist
10319 Posts
On March 15 2009 18:51 15vs1 wrote: If i get everything right the only solution for age of the second man is 60 and it is out of range. So it seems there is something wrong with conditions. I have more interesting math question. We have an n-digit number aaaa.....aaa (it can be 2222, 5555555 etc). The question is how many digits a will appear in all numbers from 1 to our number aaaa.....aaa. Example: Lets say a = 1 and n=2. In this case our number is 11. All number from 1 to 11 are: 1 2 3 4 5 6 7 8 9 10 11. There are four digits 1 (1, 10 and 11). Of course answer should be expressed in terms of a and n. I fixed it to 30-60 later and you are right. 0-9 a = 1 0-99 a = (9 + 10) 0-999 a = (9 + 10) x 9 + 100 0-9999 a = ([9 + 10] x 9 + 100) x 9 + 1000 = (19 x 9 + 100) x 9 + 1000 a = 10^(n-1) + 9 x (10^(n-2) + 9 x (10^(n-3) + 9 x (10^(n-4) + 9 x ... + 9 x (10^(n-(n-1) + 9 x (10^(n-n) + 9) a = 10^(n-1) + 9 x 10^(n-2) + 9^2 x 10^(n-3) + 9^3 x 10^(n-4) + .... + 9^(n-(n-1)) x 10^(n-(n-1)) + 9^(n-n) x 10^(n-n) | ||
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