DJ Mewdeon ft Dan Punk Posted February 14, 2015 Share Posted February 14, 2015 On 2/14/2015 at 12:44 AM, nickcrash said: RIDDLE I have millions of yes, yet I live in darkness. I have millions of ears, yet only four lobes. I have no muscle, yet I rule over two hemispheres. What am I? .... I was so confused by this riddle until I read the answer that someone else posted and realized that it was supposed to be written such that the subject had millions of eyes, not millions of yes. Quote Link to comment Share on other sites More sharing options...
Dhanush D Bhatt Posted February 15, 2015 Author Share Posted February 15, 2015 On 2/14/2015 at 5:51 PM, Sightless Cyber Avenger said: .... I was so confused by this riddle until I read the answer that someone else posted and realized that it was supposed to be written such that the subject had millions of eyes, not millions of yes. IKR? Even i was confused but then realised since it says lives in darkness it had to be eyes. Still no satisfactory answer to my riddle. Come on guys. It is a toughie though Quote Link to comment Share on other sites More sharing options...
NickCrash Posted February 15, 2015 Share Posted February 15, 2015 Oh damn, and I was certain I looked for typos. Sorry Dan. Good job Dhan. On 2/14/2015 at 9:16 AM, Dhanush D Bhatt said: My next riddle:A teacher wanted to check the intelligence of 3 of her students. She tells them to sit in a triangle so each one can see the other 2. She tells them to close their eyes. She tells them 'I am going to place a blue or red cap on all 3 of you. When i tell you to open your eyes, you have to raise your hands if you see a blue cap'. She places a blue cap on all 3 of them.When they open their eyes, all 3 of them raise their hands. Shortly afterwards, one boy says 'I have a blue cap'. How did he come to this conclusion? It took me a bit, but I think my method works Reveal hidden contents Alright so we have objects a,b,c and all see at least one blue hat. The person that says "I have a blue cap" is going to be named a, the one on his right b, and the one on his left, c. If all had a red hat, then nobody raises their arm. If two of them had a red hat, then 2 would raise their arm looking at the third, and 1 wouldn't. He would know he had a blue cap. Easy part out If one of them had a red hat, then all raise their arms. A (assuming he has a red cap) looks at b and c, b at c, and c at b. A wouldn't know. B and C would, because they know A doesn't have one, so what is the other one looking at?So C sees B's raised arm and knows he has a blue cap. Same for B If all of them have a blue cap, then all raise their arms, as they all see 2 people with blue caps A sees that C and B are not making the move (addressed in previous case), so he knows that the others see exactly what he does. Therefore after a short time seeing B's and C's reactions, he declares he has a blue hat. Quote Link to comment Share on other sites More sharing options...
Dhanush D Bhatt Posted February 16, 2015 Author Share Posted February 16, 2015 @nick *applauds. You got it to the detail i was expecting. Good job.Now for the killer ace i have been saving up.Ten people have a red cap or a blue cap on them randomly. The number of red or blue caps is not fixed. They are not allowed to communicate with each other IN ANY WAY. How do they arrange themselves in a straight line such that all the blue caps are separated from the red caps? Quote Link to comment Share on other sites More sharing options...
Gaunt Posted February 16, 2015 Share Posted February 16, 2015 On 2/16/2015 at 4:34 AM, Dhanush D Bhatt said: Ten people have a red cap or a blue cap on them randomly. The number of red or blue caps is not fixed. They are not allowed to communicate with each other IN ANY WAY. How do they arrange themselves in a straight line such that all the blue caps are separated from the red caps? They can't? No wait can they see? if yes it's easy Quote Link to comment Share on other sites More sharing options...
Serythe Posted February 16, 2015 Share Posted February 16, 2015 (edited) On 2/16/2015 at 4:34 AM, Dhanush D Bhatt said: @nick *applauds. You got it to the detail i was expecting. Good job. Now for the killer ace i have been saving up. Ten people have a red cap or a blue cap on them randomly. The number of red or blue caps is not fixed. They are not allowed to communicate with each other IN ANY WAY. How do they arrange themselves in a straight line such that all the blue caps are separated from the red caps? Reveal hidden contents Hmm.. kinda difficult but I'll take a guess. Perhaps there is a mirror, reflection, or some sort of way that you can see what color hat you have on? And if everyone is able to see, you would know who has what color hat on. Then, from that point, since your hat color is known as well as the others', they can arrange themselves in order according to their hat color? If not, then it's possible there is someone that's telling the people where to stand - if you think about it, it says that the ten people cannot communicate with each other - it doesn't say someone (without a cap) can't communicate with them. For example, it could be some kind of an experiment or test, etc. This would make sense in that case. On 2/16/2015 at 7:03 AM, Dhanush D Bhatt said: I am sorry no. There are no external people involved and no mirrors etc. Consider them in an empty room. This isn't my ace for nothing you know @Gaunt. They can see each other's caps. Not their own Lol, dang, it's pretty hard. Hopefully it won't be something extremely obvious. Well, I need to think some more about it in any case. Okay, by any chance, please don't tell me this is the answer; Reveal hidden contents Is it possible that they took the caps off and arranged the caps in a straight line according to what color the caps are? That way it would be all blue and then all red caps in a straight line? The riddle doesn't seem to rule this out lol so I figured I'd try it. Edited February 16, 2015 by Serythe Quote Link to comment Share on other sites More sharing options...
Dhanush D Bhatt Posted February 16, 2015 Author Share Posted February 16, 2015 (edited) On 2/16/2015 at 5:49 AM, Serythe said: Reveal hidden contents Hmm.. kinda difficult but I'll take a guess. Perhaps there is a mirror, reflection, or some sort of way that you can see what color hat you have on? And if everyone is able to see, you would know who has what color hat on. Then, from that point, since your hat color is known as well as the others', they can arrange themselves in order according to their hat color? If not, then it's possible there is someone that's telling the people where to stand - if you think about it, it says that the ten people cannot communicate with each other - it doesn't say someone (without a cap) can't communicate with them. For example, it could be some kind of an experiment or test, etc. This would make sense in that case. I am sorry no. There are no external people involved and no mirrors etc. Consider them in an empty room. This isn't my ace for nothing you know @Gaunt. They can see each other's caps. Not their own Edited February 16, 2015 by Dhanush D Bhatt Quote Link to comment Share on other sites More sharing options...
Gaunt Posted February 16, 2015 Share Posted February 16, 2015 On 2/16/2015 at 7:03 AM, Dhanush D Bhatt said: I am sorry no. There are no external people involved and no mirrors etc. Consider them in an empty room. This isn't my ace for nothing you know @Gaunt. They can see each other's caps. Not their own If they can see each other then they will try to go near someone with a blue/red hat and if that someone pushes them away their hat has the opposite color so gradually two groups will form and from there a line Quote Link to comment Share on other sites More sharing options...
Serythe Posted February 16, 2015 Share Posted February 16, 2015 On 2/16/2015 at 7:19 AM, Gaunt said: If they can see each other then they will try to go near someone with a blue/red hat and if that someone pushes them away their hat has the opposite color so gradually two groups will form and from there a line This might be possible but isn't that technically communicating in some way? I don't know lol Quote Link to comment Share on other sites More sharing options...
Dhanush D Bhatt Posted February 16, 2015 Author Share Posted February 16, 2015 On 2/16/2015 at 7:19 AM, Gaunt said: If they can see each other then they will try to go near someone with a blue/red hat and if that someone pushes them away their hat has the opposite color so gradually two groups will form and from there a line Some flaws in your solution: 1) This requires some prior planning and communication before to let everyone know of the plan. 2) The person who is pushing the other away will not know if that person has the cap of the opposite color since he doesn't know his own. Quote Link to comment Share on other sites More sharing options...
Gaunt Posted February 16, 2015 Share Posted February 16, 2015 On 2/16/2015 at 8:29 AM, Dhanush D Bhatt said: Some flaws in your solution: 1) This requires some prior planning and communication before to let everyone know of the plan. 2) The person who is pushing the other away will not know if that person has the cap of the opposite color since he doesn't know his own. oh then i'm really lost... Quote Link to comment Share on other sites More sharing options...
Serythe Posted February 16, 2015 Share Posted February 16, 2015 (edited) Hmm wait a second, don't tell me.. Reveal hidden contents Is it possible that one person would stand first, and since other people can see their hat, stand accordingly? Like, if they alternated so that it would be the first person with a red hat, then the second with a blue hat, and so on. And, since you cannot see your own hat, someone that can see it could stand in between the first person and whoever stood next to them. People could just change their position In the straight line to form an alternating pattern, which would separate the red hats from blue hats. ---unless, it needs to be where all red hats are on one side and then all blue hats are on the other - or in other words, alternating hat colors isn't allowed. ----- Some flaws in your solution:1) This requires some prior planning and communication before to let everyone know of the plan.2) The person who is pushing the other away will not know if that person has the cap of the opposite color since he doesn't know his own. Not sure if this post is too soon from my last one, however, was my guess in my last post (edited it earlier lol) at least close to the answer? I'm really curious what the answer could be, haha. ----- Lol okay, I wanted to check just to make sure. I can rule that out now, too xD --- Edited February 16, 2015 by Serythe Quote Link to comment Share on other sites More sharing options...
Dhanush D Bhatt Posted February 16, 2015 Author Share Posted February 16, 2015 (edited) On 2/16/2015 at 8:40 AM, Serythe said: Not sure if this post is too soon from my last one, however, was my guess in my last post (edited it earlier lol) at least close to the answer? I'm really curious what the answer could be, haha. Nuh-uh. The people WITH THE CAPS ON (lol i wrote this with the caps on ) have to arrange themselves in a line such that blah blah. Edited February 16, 2015 by Dhanush D Bhatt Quote Link to comment Share on other sites More sharing options...
Aurorix Posted February 16, 2015 Share Posted February 16, 2015 Have someone with no cap on tell them how to line up Quote Link to comment Share on other sites More sharing options...
NickCrash Posted February 16, 2015 Share Posted February 16, 2015 On 2/16/2015 at 4:34 AM, Dhanush D Bhatt said: Now for the killer ace i have been saving up. Ten people have a red cap or a blue cap on them randomly. The number of red or blue caps is not fixed. They are not allowed to communicate with each other IN ANY WAY. How do they arrange themselves in a straight line such that all the blue caps are separated from the red caps? Duuuude.... Reveal hidden contents R+B=10 THEREFORE EVEN NUMBA MOTHAFATHA!!!!!!!!!!!!!!!!!!!111 Ok. Because I like putting tags on people I shall name them: 0,1,2,3,4,5,6,7,8,9 One of those guys will be the observer. That's.... 0 (because if I don't solve it he'll still be worthless) Now the observer wears a hat but does a thing. He divides the others into a group of even and a group of odd That means that one group will have 2,4,6,or 8 people with the same cap colour and the other will have 1,3,5,7 or 9 with same cap. 0 doesn't know what he wears, but sees everyone else So, if he sees 1,3,5,7, or 9 people with a specific colour of hats, say red, he puts them in a row. He puts all the others in another row and himself in the middle of the two rows In this way, if he wears red, both groups are equal and he is separating from the blues If he wears blue, he stands behind all blue, and behind him are all the reds, so he still separates them. Quote Link to comment Share on other sites More sharing options...
laggless01 Posted February 16, 2015 Share Posted February 16, 2015 Or: one stands in a place wearing a hat (doesn't matter which one). The next stands next to him.Now the algorithm comes: every time, the next guy will go standing where the blue and red hats meet. If they're all the same color, the guy will just stand at an end of the row.For example:BBRRRthe next guy will go to the place between a B and R. Quote Link to comment Share on other sites More sharing options...
Dhanush D Bhatt Posted February 17, 2015 Author Share Posted February 17, 2015 @Serythe : Nope they can't stand alternating.@Nick : Hmm that works only if they are allowed to prompt each other where to go. Then it wouldn't be much of a riddle would it?@Lagless: You have my respect sir *applauds Quote Link to comment Share on other sites More sharing options...
Serythe Posted February 17, 2015 Share Posted February 17, 2015 On 2/17/2015 at 6:19 AM, Dhanush D Bhatt said: @Serythe : Nope they can't stand alternating. @Nick : Hmm that works only if they are allowed to prompt each other where to go. Then it wouldn't be much of a riddle would it? @Lagless: You have my respect sir *applauds Lol alrighty. And does that mean Lagless' answer was correct? I'm curious to know what the answer is. Quote Link to comment Share on other sites More sharing options...
NickCrash Posted February 17, 2015 Share Posted February 17, 2015 Lagless's answer is the correct one. Quote Link to comment Share on other sites More sharing options...
laggless01 Posted February 17, 2015 Share Posted February 17, 2015 I already saw (and solved) that puzzle before, so I was suprised it to be the trump.Anyway, I'll put in a riddle that took me hours to solve:There is this cycle: 37 -> 58 -> ? -> 145 -> 42 -> 20 -> 4 -> 16 -> 37What number belongs on the ?, and why? Quote Link to comment Share on other sites More sharing options...
Etesian Posted February 17, 2015 Share Posted February 17, 2015 Reveal hidden contents Square and add up the digits, so for example 37 is 3x3+7x7=58, it works on all numbers. So ? is 25+64=89. Quote Link to comment Share on other sites More sharing options...
Dhanush D Bhatt Posted February 18, 2015 Author Share Posted February 18, 2015 New riddle!A criminal is sentenced to death. The king allows him to die in a peculiar manner. The king says 'Tell me a statement. If it is True, you will be hanged. If it is false, you will be fed to lions'. Minutes later, the criminal was found to be walking free. How? Quote Link to comment Share on other sites More sharing options...
Gaunt Posted February 18, 2015 Share Posted February 18, 2015 The criminal said "i will be fed to lions" that should create a logic paradox allowing him to live Quote Link to comment Share on other sites More sharing options...
Dhanush D Bhatt Posted February 18, 2015 Author Share Posted February 18, 2015 On 2/18/2015 at 10:23 AM, Gaunt said: The criminal said "i will be fed to lions" that should create a logic paradox allowing him to live Correctumundo Quote Link to comment Share on other sites More sharing options...
laggless01 Posted February 18, 2015 Share Posted February 18, 2015 On 2/17/2015 at 9:18 PM, Etesian said: Reveal hidden contents Square and add up the digits, so for example 37 is 3x3+7x7=58, it works on all numbers. So ? is 25+64=89. Correct! Didn't expect it to be solved that quickly...or have you been playing professor Layton? Quote Link to comment Share on other sites More sharing options...
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