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| Riddle Answering; Someone asks a riddle, the one who answeres correctly has to ask the next. | |
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| Tweet Topic Started: Aug 20 2010, 11:31 AM (12,850 Views) | |
| Blique | Aug 27 2010, 05:43 PM Post #51 |
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Jack-of-All-Trades and Such
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The riddle doesn't require much complicated thinking. If I had to categorize it under a school level, it's kinda geometry-ish. |
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| Tetsuki | Aug 27 2010, 05:49 PM Post #52 |
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Ancestor
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But why doing it easy if we can get to it with a difficult answer aswell? |
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| kycse | Aug 27 2010, 07:16 PM Post #53 |
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Practitioner
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Easiest way would be calculating it... I assume we can't just tell by looking at it. However it has to be a way that can determine whether it's less than the half, more than the half or exactly the half, even if the difference is just 1ml or even less... If it were a cylinder it would be kinda easy... but since it's a cone without it's top it's a bit more difficult. But calculating the Volume of the water in it is probably more difficult. Besides we don't have tools to take the EXACT measurements... So what's easy in geometry, doesn't need exact measurements and can still calculate the thing down to an ml or even less? |
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| Blique | Aug 27 2010, 07:43 PM Post #54 |
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Jack-of-All-Trades and Such
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Not sure about the ml thing...It's not too small of a difference. Actually, I'm not sure about the bucket part now. >.>; Maybe it was a cylindrical bucket instead of a cone. Yeah, it probably was. >>;; Sorry. |
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| kycse | Aug 27 2010, 07:52 PM Post #55 |
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Practitioner
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I just said cylindrical because it's easier to calculate with: V = pi * r² * height ... But we still can't determine the water volume or the unused volume without some kind of measurement tools (except guessing) And the ml thing is just me thinking, since I love mathematics I can't think differently :> 50% = half, < 50% means less, so 49,9999999% would be less already which equals to even less then a ml... That's just a definition problem on my part. But if it's solvable through geometry, then there are given measurements, aren't there? (Since we can't measure ANYTHING correctly without tools, we can't determine ANY constant like height or radius ...) |
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| Blique | Aug 27 2010, 08:03 PM Post #56 |
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Jack-of-All-Trades and Such
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It's not really geometry, measurement-wise, aside from dealing with halves. It's more about the shape. |
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| kycse | Aug 27 2010, 08:17 PM Post #57 |
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Practitioner
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Ahhh, does it have a lid or some way to close the open side? And is it transparent? If yes, you can just lay it down on the side and look where the water stand is. |
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| Blique | Aug 27 2010, 09:40 PM Post #58 |
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Jack-of-All-Trades and Such
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Water stand? And no lid and no transparency. |
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| Tetsuki | Aug 28 2010, 03:13 AM Post #59 |
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I finally found a way and tested it. I could measure without any tolls, but it has to be a cylindrical bucket. First I thought we could try to calculate it if we incline it to a certain point but then I had a idea. Wouldn't it be that if the water in the bucket is exactly on the edge , when tipped, and at the same time the water is on the edge of the bottom of the bucket it would be exactly one half bucket geometrical speaking? And also if it's less then the bottom edge it would be less then half a bucket. I tried it out with some cylindrical things and it works! |
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| kycse | Aug 28 2010, 03:53 AM Post #60 |
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Practitioner
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Ahh, interesting thought. I hate ( "drawing" ) geometry. But it should be right, the diagonal should be that way. (Well I still wonder if it you could measure a difference of 1ml with that method, but for now it should be right) |
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| Tetsuki | Aug 28 2010, 03:56 AM Post #61 |
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If you have a hell of an eyesight and the bucket is exactly a cylinder it should be possible. |
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| Blique | Aug 28 2010, 04:56 AM Post #62 |
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Jack-of-All-Trades and Such
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Yup, that's the answer! ^o^ |
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| Tetsuki | Aug 28 2010, 06:16 AM Post #63 |
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hm. A riddle? There is a golden city and called so because it is made of gold. Therefore every crook wants to get in, but there is only one large door with a large guard, we will just call him "Hakan". Hakan asks everyone who wants to get in, for a password. Is it wrong, he will be killed and "won't get in" A spy sneaks close to this gate to listen and hears the following: A peasant comes to the gate. Hakan: "12 - what is your response?" Bauer: "6" - and he may pass. A nun comes your way. Hakan: "6 - what is your response?" Nun: "3" - and she may pass. Now the spy thinks (as some of you aswell ;)) how easy it was and goes to the gate. Hakan: "8 - what is your response?" The spy grins and says "4". Hakan kills him, that approach was wrong So what is the right answer to the last question? What is the answer to "8"? Please no guessing a real explanation^^
Edited by Tetsuki, Aug 28 2010, 06:19 AM.
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| V | Aug 28 2010, 06:27 AM Post #64 |
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Vivaciously Vexacious Vivification
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I have a couple answers. All math related but hey thats how I think. Could be 5, as in the # of letters in the number spelled out. Twelve = 6 Six = 3 eight = 5. Though I feel that isn't the answer. |
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| Tetsuki | Aug 28 2010, 06:28 AM Post #65 |
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It is^^ It was an easy riddle ![]() Next ones for you. |
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| kycse | Aug 28 2010, 03:00 PM Post #66 |
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Practitioner
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I think to abstract... 12 has 2 digits, 2 digits * 3 = 6, 6 is a digit, so 1 digit * 3 = 3, same for 8 then :> |
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| guilover | Aug 29 2010, 12:22 PM Post #67 |
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Expert
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so v, what is your riddle? |
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| Tetsuki | Sep 4 2010, 03:42 PM Post #68 |
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The I will ask a new one: I have one, you have one. If you remove the first letter, a bit remains. If you remove the second, bit still remains. After much trying, you might be able to remove the third one also, but it remains. It dies hard! |
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| Blique | Sep 4 2010, 08:38 PM Post #69 |
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Jack-of-All-Trades and Such
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Habit, I think! 8D |
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| Tetsuki | Sep 4 2010, 09:25 PM Post #70 |
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yup, right answer. Your turn |
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| Blique | Sep 6 2010, 07:29 AM Post #71 |
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Jack-of-All-Trades and Such
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Humm... Two children were born in different years, in different months, and on different days, yet they are twins. How is this possible? |
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| guilover | Sep 6 2010, 03:54 PM Post #72 |
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one was born just before midnight on december 31, the other right after midnight on january 1. |
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| Blique | Sep 6 2010, 05:35 PM Post #73 |
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Jack-of-All-Trades and Such
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Yup~ |
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| guilover | Sep 7 2010, 12:57 AM Post #74 |
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Expert
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my eyes cannot see, my skin is thick, i live and grow under the earth. some people with red hair seem to have quite a love/hate relationship with me. what am i? |
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| amginE | Sep 7 2010, 08:40 PM Post #75 |
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what's taters? POTATOES. boil em, mash em, stick em in a stew? (i'm either spot on or way off...) |
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| apoptoxin4869 | Sep 8 2010, 07:24 PM Post #76 |
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Ruler of the flying squirrels
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earthworm? |
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| guilover | Sep 8 2010, 07:40 PM Post #77 |
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amg got it! i made that one up on the spot... you like?
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| Blique | Sep 8 2010, 10:35 PM Post #78 |
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Jack-of-All-Trades and Such
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Ehh, but potatoes don't have thick skin...? It's rather thin. |
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| kycse | Sep 9 2010, 11:08 AM Post #79 |
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Practitioner
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And why are they hated/loved by some red haired people? |
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| amginE | Sep 9 2010, 05:11 PM Post #80 |
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yeah, the thick skin part made me doubt my answer the most. im guessing red haired people -> the irish. and as you know, the irish eat lots of potatoes. they either love it or hate it from eating so much. New riddle: you are in a band of five robbers who found 51 pieces of gold. it is decided that the money will be split up in this manner: each robber draws straws. the one with the longest straw gets to propose the split (e.g. i get 11, you get 10, he gets 9, etc.) then all the robbers, including the one with the proposal, votes 'yes' or 'no' to the split. if 50% or more of them vote 'yes', then it is approved and the gold is split in that manner. if not, the robber who proposed it gets killed and the robber with the next longest straw gets to propose until a proposal passes. now, you draw the longest straw. you know the following two facts. a) all the other robbers are super geniuses (and you are too, right?) b) the order of importance for all the other robbers are as follows: 1) stay alive 2) get as much gold as possible 3) kill off as many other robbers as possible so for example, if a robber gets the choice of either 5 gold with 2 other robbers dead and 6 gold with 1 other robber dead, he would choose the second option. if he gets the choice of 5 gold with 2 other robbers dead and 5 gold with 1 other robber dead, he would choose the first option. what should be your proposal? you: ? gold robber with second longest straw: ? gold robber with third longest straw: ? gold robber with fourth longest straw: ? gold robber with shortest straw: ? gold If you heard this one before, then do NOT answer
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| V | Sep 9 2010, 05:49 PM Post #81 |
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Vivaciously Vexacious Vivification
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What is the goal... for you to get the most and not die? |
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| amginE | Sep 9 2010, 06:14 PM Post #82 |
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i sure hope that is your goal. unless of course you're suicidal or decided to be a robber for the fun of it.
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| Blique | Sep 9 2010, 09:18 PM Post #83 |
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Jack-of-All-Trades and Such
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Hmmm... Me: 17 gold Robber with second longest straw: 0 gold Robber with third longest straw: 0 gold Robber with fourth longest straw: 17 gold Robber with shortest straw: 17 gold Though, that doesn't seem like a fulfilling answer. Hmm... |
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| apoptoxin4869 | Sep 9 2010, 11:47 PM Post #84 |
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Ruler of the flying squirrels
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poop |
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| kycse | Sep 10 2010, 12:03 PM Post #85 |
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Practitioner
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(Irish people have red hair? I wonder..., perhaps I should learn more about people or you are pretty biased) Isn't there a mistake in the riddle. Your example's don't make sense at any rate. Since I am the robber who is the first to propose a split, I am the first (and only one) to be killed in the first round. If 3 robbers (excluding myself) vote against the split, I'll be killed ==> you have t satisfy two robbers apart from yourself If 3 robbers (including myself) vote for the split, the split is done ==> no one is killed. So, HOW can any of the others be killed when i go first? So there has to be some rule that robbers who disagree are killed, else your examples don't make sense. So I can't vote later anyway... Although there's quite a contradiction... (At least I can't think of it any other way) Now then, let's do this backwards: If we have 2 people, one of them has to vote "yes" (since it's >= 50%) ==> #5 = 0, #4 = 51, because #4 can vote yes and therefore the split passes, So #4 will always vote no if no one gives him 51. If we have 3 people, 2 have to vote yes: Maximum Split for person 3: #5 = 1, # 4 = 0, #3 = 50, because #5 doesn't has any chance to get even one coin if there are only two people ==> one coin will always satisfy him. (theoretically) So #3 will always say no, unless someone offers him 50 coins. If we have 4 people, 2 have to vote yes. Just like the 3 people: #5 = 1, #4 = 0, #3 = 0, #2= 50. Same reasoning as with 3 people... #4 votes no unless he/she gets 50 coins. Now we have the maximum values for each person: #5 = 0 #4 = 51 #3 = 50 #2 = 50 Dilemma: With 5 people we need 3 people to vote yes. Including yourself you need 2 votes. Seeing the earlier calculated maximum values you need to satisfy two persons. Now there are two possibilities for more or less bloodshed... If people are only satisfied by their maximum number + 1, you'll ALWAYS die, since there isn't any combination that's less or equal to 51 ( 0 + 1 + 50 + 1 = 52 > 51). Pointless struggle, just give up. If they are satisfied by their maximum number. Well the exception is #5 who'll be satisfied when he gets 1 coin. (No one is satisfied by 0 coins) It would be #1 and #4 and ( #2 or #3) get all 0. #5 gets 1. And either #2 or #3 get 50 coins. That way you'll stay alive, but have no reward. ANY other way, you'll die, at least if I understand this correctly and calculated the maximum splits correctly. Therefore my answer is: #5 = 1 #4 = 0 #3 = 0 /50 #2 = 50/0 #1 = 0 (you) Reward 0, Everyone should be alive. But no, there's something else to be considered. #4 gets no reward either way. Since no one of the others would give him a coin, so he should be satisfied by 1 coin as well: #5 = 1 #4 = 1 #3 = 0 #2 = 0 #1 = 49 Now then... I'll let this answer be for now, since I don't want to explain further branches like "if #2 promises #4 more then you, then", because that would make this whole thing pretty tricky. Edit, revised theory, my current answer: Starting back at 2 people: #5 = 0 #4 = 51 ==> #5 is satisfied by 1 coin #5 = 1 #4 = 0 #3 = 50 ==> #4 is satisfied by 1 coin, #5 by 2 coins #5 = 0 #4 = 1 #3 = 0 #2 = 50 ==> #5 is satisfied by 1 coin, #3 is satisfied by 1 coins, #4 by 2 *sigh* Seems I stopped thinking back there, so this is another possible answer, depending on the rules... #5 = 1 #4 = 0 #3 = 1 #2 = 0 #1 = 49 *Sigh* To much backthinking... Edited by kycse, Sep 10 2010, 12:25 PM.
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| amginE | Sep 10 2010, 05:49 PM Post #86 |
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Kycse, your revised answer is correct! You get to keep 49 gold by just doling off one gold each to two critical people. I'm not quite sure what you were confused about at first, so I'll go ahead and explain again. Working backwards is the right idea. For simplicity, let's call the robbers A, B, C, D, E, where A is you and E is the guy with the shortest straw. If somehow only D and E are left, D will definitely propose 51/0 split and get away with it since he himself can vote yes. With C, D, and E left, there is absolutely no way for C to pursuade D to not vote no, since even if C gives all 51 gold to D, D will still vote no, kill him off, and propose the 51/0 split as described earlier. (Remember rule #3, if amount of gold is the same, then they want to kill as many people off as possible). But if C gives E just 1 gold, E will be better off than in the situation with just D and E earlier. So C proposes 50/0/1 and the vote will pass. If B is alive as well, with 4 people, he would only need to convince one other person to vote yes, and he would be good. Looking at the result with just three people, you see that robber D gets 0 gold. So if he gives robber D 1 gold, D would definitely vote yes for him. So B proposes 50/0/1/0 and the vote will pass. Now, take a look at the result if you died. All you need to do is convince other two people. C and E are the ones who would get 0 if your vote fails. So give them each a single gold, leave yourself with 49, and your vote will pass. Thus, 49/0/1/0/1 is the answer. If you still don't understand, or think someone doesn't make sense, feel free to ask. Oh, and as for the red hair thing, I wasn't the one who came up with the riddle. It's just my interpretation. |
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| Blique | Sep 11 2010, 02:23 AM Post #87 |
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Jack-of-All-Trades and Such
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The stereotype of Irish people is that they have red hair. Maybe it's an American stereotype of them...? |
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| kycse | Sep 11 2010, 02:37 AM Post #88 |
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No, it's alright. I think I understand how you meant the examples... They were not for you, but for the xth splits. It's just, because your are first, you can either end up dead or get your proposed split. You can't kill off any of the other robbers, that's why the example confused me. (As for my answer... I confused myself by thinking about other theories. Eventually I added the most convincing one.) Anyways again something with numbers: In a mint are 10 bags of coins. However someone switched the coins in x > -1 of the bags. The real coins weight 1g / piece, the wrong coins weight 0,9g. How can you find out by weighing them only once, which bag(s) hold(s) wrong coins? (Sorry for the math, but prose isn't that good at explaining a mathematical riddle.) Edit: So it is a stereotype... wow, I have less stereotypes then I thought I did
Edited by kycse, Sep 11 2010, 02:38 AM.
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| apoptoxin4869 | Sep 11 2010, 07:34 AM Post #89 |
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Ruler of the flying squirrels
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pewp |
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| amginE | Sep 11 2010, 09:40 PM Post #90 |
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hey ky, could you elaborate on what "x > -1" means? do you mean that at least one bag has the lighter coins, but it could also be several bags? also, is it a weigh or a scale (does it give you the exact weight or does it just tell you which side is heavier)? and how many coins are in each bag? if i'm understanding it right, i think i have a solution, but it would require a weigh, not a scale, and that there be at least 500 something coins in at least one bag. Edited by amginE, Sep 11 2010, 09:50 PM.
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| guilover | Sep 12 2010, 02:26 AM Post #91 |
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blique, it is an American stereotype because we are brought up seeing commercials for lucky charms... which happens to have a leprechaun as the mascot/main advertising symbol, and everyone knows leprechauns are Irish. do i realize this is inaccurate and very biased? yes, yes i do. i am 1/2 Irish and very much NOT a ginger. i do, however, sometimes wish i were. gingers are awesome. ![]() and the potatoes you use for making baked potatoes have thick skins. the potatoes you use for mashing do not. |
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| Blique | Sep 12 2010, 06:47 AM Post #92 |
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Jack-of-All-Trades and Such
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I've never seen a potato with a thick skin... Eh? Doesn't the Lucky Charms guy have brown hair? *checks* Ooo, it is red. But "American leprechauns" in general have red (it's more orange-ish, technically) hair, from what I've seen. I even have a plushie of one. =P I've seen them in cartoons as I was growing up as well. Ah, found a redhead fact: "The highest percentage of natural redheads in the world is in Scotland (13%), followed closely by Ireland with 10%. In the US, about 2% of the population are natural redheads." Red hair is a distinguishing feature that's not as common in America as in Ireland, so it interests Americans. People like to emphasize unique traits, so they do. owo |
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| kycse | Sep 12 2010, 09:28 AM Post #93 |
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Ok, elaboration: It means there could be a bag with different coins, it might be one, maybe two, perhaps 3 etc., but it's a possibility, that no bag has switched coins too. ( x > -1 even) Each bag holds 1000 coins And you get the exact value: Like 100 coins of each bag (when no bag is switched) results 1000,0g |
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| amginE | Sep 12 2010, 11:17 AM Post #94 |
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Hey ky, thanks for the elaboration. Here's my answer. Label the bags from "A" to "J". Take 1 coin from bag A, 2 coins from bag B, 4 coins from bag C, 8 coins from bag D, 16 coins from bag E, 32 coins from bag F, 64 coins from bag G, 128 coins from bag H, 256 coins from bag I, 512 coins from bag J. The total of these coins, if none are fake, should be 1023 grams (sum up those numbers, or if you're smart, it's just 10^11 - 1). If they're all fake, it should be 920.7 grams (1023 * 0.9) Take the actual weight of the set of coins and subtract it from 1023 grams. If it's off by 0.1 grams, then it's only bag A that is fake. If it's off by 0.2 grams, then it's only bag B. If it's off by 0.4 grams, then only bag C, and so on. Now here's the beauty of using powers of two. If we had used 1 coin, 2 coins, 3 coins, 4 coins, etc. then, say, if it's off by 0.3 grams, it could either mean it's from bag C alone, or from bags A and B together. But with this combo, every different combination of bags (from no bags being fake to all bags being fake) will give a different weight. Using the above example, if it's bag C alone, then it'll be 0.4 grams off, but it's a combination of bags A and B, then it would be 0.3 grams instead. Tada. That's the mathematical solution. Now, the practical solution (no need to take and sort out bags). Weigh a bag. See if it's real (1 kg) or fake (.9 kg). The difference should be noticeable, so just compare the rest of the bags by holding them, one bag in each hand. If the first bag is real, then fake bags will be lighter, real bags will be the same. If the first bag is fake, then fake bags will be the same, real bags will be heavier. Now, the lazy solution. Send a sample coin from each bag to a chemist. Mwahahaha. |
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| kycse | Sep 12 2010, 12:21 PM Post #95 |
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Err. basically yes, just a little mistake in your explanation: 1023 is not 10¹¹ -1 (that would be 9999999999), it's 2¹⁰ - 1 (binary system and kibi, for more information: http://en.wikipedia.org/wiki/Binary_prefix#Specific_units_of_IEC_60027-2_A.2_and_ISO.2FIEC_80000 ) But yes, your answer is correct. You take 2⁰ coin(s) from the first bag, then 2¹ coin(s) from the second, 2² coin(s) from the third and so on. ( anything [including 0] power 0 is always 1, just in case someone doesn't know) The difference amgine talked about is always definite: multiply it by ten (so you have an integer, and write it as a binary number, the read it from right to left and each 1 represents a fake bag: 0000000011 would mean that bag A and B were wrong and the difference was 0.3g) Anyways, amgine your turn. |
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| guilover | Sep 12 2010, 01:40 PM Post #96 |
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these riddles are getting too difficult. they're like math extra credit...
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| apoptoxin4869 | Sep 12 2010, 04:46 PM Post #97 |
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Ruler of the flying squirrels
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agreed... i have enough calc hw to go through everyday |
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| amginE | Sep 12 2010, 05:24 PM Post #98 |
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Yeah, 2^11-1, that's what I was thinking... Stupid fingers. Okay, a no-number-crunching-involved riddle then for all you math-haters. There were nine wise men who were captured by a sadistic general. The general devises a game to kill them off. He lines them all up in a row facing the same direction, and puts on each of them either cat ears or bunny ears (the exact number of each is unknown. Not necessarily 5 and 5). This is done in a way such that each person can see the headgear of everyone else' in front of them, but not their own and not the ones behind. Then, the general goes to the person in the very back, and asks "What kind of ears do you have?" They can only reply "cat" or "bunny". If that person answers correctly, they live. If not, that person dies. The general then continues up the line until asking the question to all of them. They can't make any hidden signals or break any rules, and if they do, they all die. Fortunately, the wise men overhear about this plot, and they devised a counter-plan to save as many of them as possible. What is their plan, and how many people can be saved with certainty? |
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| Shiki | Sep 12 2010, 06:40 PM Post #99 |
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My hovercraft is full of eels.
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You know, all this bias about Irish being red-haired... I though it was the Scottish that were supposed to be seen as ginger XD Though there's a general British [or at least English] thing against ginger haired people >.>;; I don't really get it myself, but best not to mention it... Ah, I missed the math riddle -dies- Word ones are a nightmare, all I can think of after my general confusion and re-reading of around a dozen times is 'shadows', but that's no good =/ It's probably some thing to do with how they answer. I await for someone to solve this one. |
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| apoptoxin4869 | Sep 12 2010, 10:25 PM Post #100 |
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Ruler of the flying squirrels
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i actually just thought the riddle was too long and not wanted to waste time on it so i really didnt read it LOL i guess im just lazy |
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8:34 PM Jul 10
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![]](http://z4.ifrm.com/static/1/pip_r.png)



a real explanation^^

i made that one up on the spot... you like?


8:34 PM Jul 10