tag:blogger.com,1999:blog-11477841819014278602014-07-05T17:29:31.578-07:00The Weekly RiddleWas at one time weekly, but now riddles will be added sporadically.Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.comBlogger47125tag:blogger.com,1999:blog-1147784181901427860.post-54841296921451585522010-09-22T06:47:00.000-07:002010-09-29T06:38:27.775-07:00The Cosmic Number<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/_KufTlvMetK4/TJoIzwHnFOI/AAAAAAAAAYA/RzAQBf2IemY/s1600/number-4.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" px="true" src="http://3.bp.blogspot.com/_KufTlvMetK4/TJoIzwHnFOI/AAAAAAAAAYA/RzAQBf2IemY/s320/number-4.jpg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">four is cosmic</td></tr></tbody></table>4 is cosmic.<br /><br />1 is 3, 3 is 5, 5 is 4, and 4 is cosmic.<br />8 is 5, 5 is 4, and 4 is cosmic.<br />2 is 3, 3 is 5, 5 is 4, and 4 is cosmic.<br />15 is 7, 7 is 5, 5 is 4, and 4 is cosmic<br />9 is 4 and 4 is cosmic.<br />18 is 8, 8 is 5, 5 is 4, and 4 is cosmic.<br />17 is 9, 9 is 4, and 4 is cosmic. <br />100 is 10, 10 is 3, 3 is 5, 5 is 4, and 4 is cosmic.<br /><br />If you think you recognize the pattern, submit an example. If you need more examples, I can give you more.<br /><br /><a name='more'></a><br /><br /><br /><br /><br />Solution:<br />one is 3, three is 5, five is 4, and 4 is cosmic. The number of letters in each number leads to the next number in the sequence - one has 3 letters, three has 5 letters, five has 4 letters. Four is cosmic because its value equals its number of letters. I'm pretty sure it's the only number with this property. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com22tag:blogger.com,1999:blog-1147784181901427860.post-4674152709582089572010-09-16T09:06:00.000-07:002010-09-16T09:06:17.967-07:00Running Out of RiddlesWell I thought I could make it a whole year providing a new riddle every week, but it appears I am almost out. I will continue posting puzzles as I find them, but I can't guarantee a weekly riddle any more at this point. For now, an old one my dad once told me:<br /><br />What did the left eye say to the right eye?<br /><br /><br /><a name='more'></a><br /><br /><br /><br /><br />Just between you and me, something smells. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.comtag:blogger.com,1999:blog-1147784181901427860.post-14355015958960978432010-09-07T21:44:00.000-07:002010-09-16T09:07:39.546-07:00The Hateful Neighbors<div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_KufTlvMetK4/TIcUOefOCjI/AAAAAAAAAXs/kIL7mtcvL-s/s1600/Slide1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="300" src="http://1.bp.blogspot.com/_KufTlvMetK4/TIcUOefOCjI/AAAAAAAAAXs/kIL7mtcvL-s/s400/Slide1.gif" width="400" /></a></div><br />The Boggis, Bunce, and Bean families were once united in their pursuit of a common enemy, but have since developed a bitter and irreconcilable hatred for one another. Each family lives in its own house in its own part of town, and all is well and good, so long as members of different families don't cross paths - if they do, they will start brawling until the poor sheriff has to come out and break them up. They are very civil indoors, however. <br /><br />Each family needs to be able to access the post office, the general store, and the sheriff's office without encountering members of other families.<br /><br />So the sheriff has come up with a great plan; draw up plans for each family to have its own three paths, traveling from each home to a each of the three municipal buildings. In this way, for example, the Boggis family has three paths, where each path leads from its front door to the front doors of the post office, general store, and sheriff's office. Can he do this without letting the paths cross, and without digging any tunnels or building any bridges - in other words, working in a two-dimensional plane?<br /><br /><br /><a name='more'></a><br /><br /><br /><br />Solution: Unfortunately, there is none. There is no way to create three sets of paths to three locations without them crossing or meeting. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.comtag:blogger.com,1999:blog-1147784181901427860.post-14892859289595312602010-08-25T12:06:00.000-07:002010-09-22T09:48:32.485-07:001000 Bottles of Wine<div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_KufTlvMetK4/THVpQazQKII/AAAAAAAAAXY/S4BKAUEDua8/s1600/bottles+of+wine.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" ox="true" src="http://1.bp.blogspot.com/_KufTlvMetK4/THVpQazQKII/AAAAAAAAAXY/S4BKAUEDua8/s320/bottles+of+wine.jpg" /></a></div><br /><em>Borrowed from <a href="http://folj.com/">folj.com</a>. </em><br />You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned. <br />The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.<br /><br />You have over a thousand paid caterers to help with the testing and just under 24 hours to determine which single bottle is poisoned. <br /><br />You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.<br /><br />What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours?<br /><a name='more'></a><br /><br /><br />Solution directly from folj.com<br />Answer: 10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.<br /><br /><br /><br /><br />Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.<br /><br /><br /><table border="1" cellpadding="0" cellspacing="0" class="solutions"><><>lt;><><> </></></></></><tbody><tr><><><><><> </></></></></><td align="right"></td><><><><><> </></></></></><td align="center">Bottle 1</td><><><><><> </></></></></><td align="center">Bottle 2</td><><><><><> </></></></></><td align="center">Bottle 3</td><><><><><> </></></></></><td align="center">Bottle 4</td><><><><><> </></></></></><td align="center">Bottle 5</td><><><><><> </></></></></><td align="center">Bottle 6</td><><><><><> </></></></></><td align="center">Bottle 7</td><><><><><> </></></></></><td align="center">Bottle 8</td><><><><><> </></></></></></tr><><><><><> </></></></></><tr><><><><><> </></></></></><td align="right" nowrap="nowrap">Prisoner A</td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></></tr><><><><><> </></></></></><tr><><><><><> </></></></></><td align="right" nowrap="nowrap">Prisoner B</td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></></tr><><><><><> </></></></></><tr><><><><><> </></></></></><td align="right" nowrap="nowrap">Prisoner C</td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center"></td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></><td align="center">X</td><><><><><> </></></></></></tr><><><><><> </></></></></></tbody></table><br /><br /><br /><br />Here is how you would find one poisoned bottle out of eight total bottles of wine. <br /><br /><br />In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.<br /><br />With ten people there are 1024 unique combinations so you could test up to 1024 bottles of wine.<br /><br />Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a millilitre from each bottle, they will only consume the equivalent of about one bottle of wine each. <br /><br />Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combinations where all but one prisoner must sip from the wine. By avoiding these two types of combinations you can ensure no more than 8 prisoners die.<br /><br />One viewer felt that this solution was in flagrant contempt of restaurant etiquette. The emperor paid for this wine, so there should be no need to prove to the guests that wine is the same as the label. I am not even sure if ancient wine even came with labels affixed. However, it is true that after leaving the wine open for a day, that this medieval wine will taste more like vinegar than it ever did. C'est la vie.Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com12tag:blogger.com,1999:blog-1147784181901427860.post-78718381230785416992010-08-18T00:00:00.000-07:002010-09-16T09:11:42.397-07:00SEND MORE MONEYOne time in college I emailed my dad asking for money and he said I could have some if I solved the following equation:<br /><br /> SEND<br />+MORE<br />_______<br />MONEY<br /><br />Each letter represents its own digit (0-9) and multiple occurrences of the same letter represent the same digit (eg if one of the E's represents a 3, they all do).<br /><br /><a name='more'></a><br /><br /><br /><br />Solution:<br /> 9567<br />+ 1085<br />-----------<br />10652Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com13tag:blogger.com,1999:blog-1147784181901427860.post-12311598732709414022010-08-13T08:45:00.000-07:002010-08-25T12:42:54.740-07:00A Boat in a Tank<div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_KufTlvMetK4/TGVoZ5y_ojI/AAAAAAAAAXA/tLVUWNbcUWQ/s1600/rowboat.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" ox="true" src="http://2.bp.blogspot.com/_KufTlvMetK4/TGVoZ5y_ojI/AAAAAAAAAXA/tLVUWNbcUWQ/s320/rowboat.gif" /></a></div>Imagine you are in a small rowboat floating in a swimming pool. There's a big rock in the boat and you drop it overboard. Does the water level rise or fall? Why?<br /><br /><a name='more'></a><br /><br /><br /><br /><br />Solution:<br />The water level will go down. Imagine two rocks, the same size and shape, the only difference being that one of them is only slightly more dense than water, and the other is really really dense, made of lead. When dropped into the pool, each rock will displace the same amount of water, because they have the same volume. When something sinks, it displaces a quantity of water equal to its volume. But when placed in a boat, the rocks are adding weight to a floating object - the boat. Floating objects displace a quantity of water equal to their <em>weight.</em> In the boat, the different rocks will have different effects. The light rock won't make much difference - the boat will displace a bit more water than it would otherwise, making the water level rise slightly. But the heavy rock will really weigh the boat down, causing it to displace a lot of water, and causing the water level to rise a lot. <br /><br />So to answer the question of what would happen if a rock was thrown off a boat, we need to compare displacement when the rock is in the boat (floating) to when the rock is at the bottom of the pool (sunk). Since a rock is more dense than water, it has more weight than volume, relative to water.So more water would be displaced by its weigh than by its volume. Therefore more water is displaced when it's in the boat than at the bottom of the pool; so when it's dropped overboard, less water is displaced and the water level goes down. <br /><br />The opposite would happen if you were chained to the bottom and pulled an empty two-liter bottle under the water. First, it's only displacing water equal to its weight, because it's floating. But then when you pull it under water, it's displacing water equal to its volume, which is much more. Then the water level rises. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com3tag:blogger.com,1999:blog-1147784181901427860.post-15452379187278671092010-08-04T21:31:00.000-07:002010-08-12T11:50:07.231-07:00The Magic Square<div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/_KufTlvMetK4/TFrEEkmh3NI/AAAAAAAAAWY/AGvvP9HBgPQ/s1600/Magic+Square+problem.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" bx="true" src="http://4.bp.blogspot.com/_KufTlvMetK4/TFrEEkmh3NI/AAAAAAAAAWY/AGvvP9HBgPQ/s320/Magic+Square+problem.png" /></a></div><br />This is one I come back to when I'm bored and all I have is pen and paper. I also read in a biography of Benjamin Franklin that he used to do this when he was stuck in boring meetings. Construct a 3x3 grid of numbers, using numbers 1 through 9, and arrange the numbers in the square such that every row, column, and diagonal (diagonals through the center) adds up to 15. <br /><br /><br />Too easy? Now construct a 4x4 grid made out of numbers 1 through 16, such that every row, column, and diagonal adds up to 34. This one is killing me because I figured it out once, but can't seem to rediscover the solution. There are ways to do this for grids 5x5, 6x6, and up, though they no doubt get very difficult.<br /><br /><a name='more'></a><br /><br /><br /><br /><br />One solution to the 3x3 square is below. There are other solutions that are simply rotations of this one - can you find a solution that is not a rotation of the square below? As for the 4x4 square, I'm still trying to figure it out. <br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/_KufTlvMetK4/TGRCPLrxZZI/AAAAAAAAAWo/mzTI-tPAdAs/s1600/magic+square.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" ox="true" src="http://4.bp.blogspot.com/_KufTlvMetK4/TGRCPLrxZZI/AAAAAAAAAWo/mzTI-tPAdAs/s320/magic+square.png" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div>Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com5tag:blogger.com,1999:blog-1147784181901427860.post-17404368521359058122010-07-28T14:17:00.001-07:002010-07-28T14:17:48.439-07:00No Riddle This WeekI can't think of any and I'm too busy with work, sorry everybody. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.comtag:blogger.com,1999:blog-1147784181901427860.post-53533288941223475252010-07-21T00:00:00.000-07:002010-07-21T00:00:01.741-07:004=5The other day, someone in our creative services department tried convincing me that 4=5. He offered a convincing proof, which is displayed below. But what is wrong with it? <br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_KufTlvMetK4/TD2_wCp2UaI/AAAAAAAAAV8/xXtBJ-mhI-g/s1600/4%3D5+Proof.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="400" rw="true" src="http://2.bp.blogspot.com/_KufTlvMetK4/TD2_wCp2UaI/AAAAAAAAAV8/xXtBJ-mhI-g/s400/4%3D5+Proof.jpg" width="372" /></a></div>Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com2tag:blogger.com,1999:blog-1147784181901427860.post-12414519519112715202010-07-14T00:00:00.000-07:002010-07-14T00:00:05.677-07:00Pirates, Part 2Again, taken from <a href="http://www.folj.com/puzzles/difficult-logic-problems.htm">folj.com</a>. <br /><div class="separator" style="clear: both; text-align: center;"><a href="http://cdn2.ioffer.com/img/item/906/421/91/ZJSWRCJLVu52Pyd.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" rw="true" src="http://cdn2.ioffer.com/img/item/906/421/91/ZJSWRCJLVu52Pyd.jpg" width="320" /></a></div><br /><br />The five pirates mentioned previously are joined by a sixth, then plunder a ship with only one gold coin. <br /><br /><br />After venting some of their frustration by killing all on board the ship, they now need to divvy up the one coin. They are so angry, they now value in priority order: <br /><br />1. Their lives<br />2. Getting money<br />3. Seeing other pirates die.<br /><br /><br />So if given the choice between two outcomes, in which they get the same amount of money, they'd choose the outcome where they get to see more of the other pirates die. <br /><br /><br />How can the captain save his skin?Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com2tag:blogger.com,1999:blog-1147784181901427860.post-49278716899726800692010-07-07T13:18:00.000-07:002010-07-07T13:30:42.085-07:00The Blind Date Bachelor<div class="separator" style="clear: both; text-align: center;"><a href="http://listsoplenty.com/blog/wp-content/uploads/2009/12/the-bachelor.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" rw="true" src="http://listsoplenty.com/blog/wp-content/uploads/2009/12/the-bachelor.jpg" width="254" /></a></div><em>(Modified from a puzzle about a sultan and his harem I heard from Jon Huang.)</em> The Blind Date Bachelor is the newest dating show, in which where there is 1 bachelor and 4 contestants trying to win his heart. He will meet each contestant for the first time on a blind date. At the end of the date, he must choose whether to marry her or never see her again. If he marries her, the game is over. If he rejects her, he is set up on a date with the next contestant and repeats the process. If he rejects the first 3, he marries the last one automatically. He is able to compare and rank contestants that he has already met, but will not know for sure who he likes best until he has met them all. It's very important that he marry the best one, or he will spend the rest of his life wondering what could have been. What strategy will maximize his chances of finding the best mate-for-life?<br /><br />3 Bonus Questions: What is the probability of winning using the best strategy? What if there are 5 contestants, not 4? And finally, what if there are <em>n</em> contestants? Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com1tag:blogger.com,1999:blog-1147784181901427860.post-62003246908273323022010-06-30T07:30:00.000-07:002010-07-08T11:11:20.327-07:00PiratesI've seen this one before and haven't solved it. Go ahead and comment answers in the blog, but not on Buzz or else people will see them. I quote this one from <a href="http://www.folj.com/puzzles/difficult-logic-problems.htm">folj.com</a>, but I've seen it before in other places as well. <br /><div class="separator" style="clear: both; text-align: center;"><a href="http://static.howstuffworks.com/gif/pirate-9.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://static.howstuffworks.com/gif/pirate-9.jpg" width="254" /></a></div><br /><br /><div class="separator" style="clear: both; text-align: center;"></div><br />Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).<br /><br />The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go "Aye", the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side. <br />If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain. <br /><br />What is the maximum number of coins the captain can keep without risking his life?<br /><br /><br /><a name='more'></a><br /><br /><br />Solution: See the "Solutions" page: link is at the top of the sidebar on the right. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com9tag:blogger.com,1999:blog-1147784181901427860.post-83355987790650813332010-06-23T07:50:00.000-07:002010-07-01T03:03:00.487-07:00Dots and LinesThis is a very old one, and many of you may have seen it before. I try to avoid posting puzzles with outside-the-box solutions, but I made an exception for this one. Using 4 straight, connected lines, connect all 9 dots.<br /><br /><br /><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_KufTlvMetK4/TCIdK0JKTOI/AAAAAAAAAVo/PKeMoX2E0jQ/s1600/Dots+and+Lines.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="185" ru="true" src="http://2.bp.blogspot.com/_KufTlvMetK4/TCIdK0JKTOI/AAAAAAAAAVo/PKeMoX2E0jQ/s200/Dots+and+Lines.png" width="200" /></a></div><br /><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><br /><a name='more'></a><br /><br /><br /><br /><br /><br />Solution:<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/_KufTlvMetK4/TCtVaJvqvuI/AAAAAAAAAV0/nU93LKm8NFU/s1600/Dots+and+Lines+Solution.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="172" ru="true" src="http://4.bp.blogspot.com/_KufTlvMetK4/TCtVaJvqvuI/AAAAAAAAAV0/nU93LKm8NFU/s200/Dots+and+Lines+Solution.png" width="200" /></a></div><br /><br /><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"> </div>Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com0tag:blogger.com,1999:blog-1147784181901427860.post-63268587168279772412010-06-16T07:28:00.000-07:002010-06-23T08:09:57.821-07:00Heads and TailsThis is borrowed from <a href="http://www.folj.com/puzzles/difficult-logic-problems.htm">folj.com</a> and I don't know the solution. Submit ideas in the comments section - maybe you'll get it before I do. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://allrarecoins.com/im/2009/08/rare-coins-1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="166" qu="true" src="http://allrarecoins.com/im/2009/08/rare-coins-1.jpg" width="200" /></a></div><br />There are twenty coins sitting on the table, ten are currently heads and ten are currently tails. You are sitting at the table with a blindfold and gloves on. You are able to feel where the coins are, but are unable to see or feel if they are heads or tails. You must create two sets of coins. Each set must have the same number of heads and tails as the other group. You can only move or flip the coins, you are unable to determine their current state. How do you create two even groups of coins with the same number of heads and tails in each group? [Note: I assume you need a guaranteed method of getting even groups, rather than method that will likely work.]<br /><br /><a name='more'></a><br /><br /><br /><br /><br /><br />Solution:<br /><br />Rob got the solution faster than I could, and here it is:<br />Split into two groups of 10. Group A will have X heads, and 10-X tails. Group B will have X tails, and 10-X heads. Flip every coin in group B. Both groups now have X heads and 10-X tails.Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com4tag:blogger.com,1999:blog-1147784181901427860.post-39898230063364336722010-06-02T00:00:00.000-07:002010-06-16T06:59:32.442-07:00The Lantern and the Bridge<div class="separator" style="clear: both; text-align: center;"><a href="http://www.engineeringdaily.net/wp-content/uploads/2009/08/rope-bridge.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" gu="true" height="320" src="http://www.engineeringdaily.net/wp-content/uploads/2009/08/rope-bridge.jpg" width="245" /></a></div>A small family is being pursued by an unknown enemy in the middle of a dark, dark night. It comes to a deep chasm spanned by a narrow bridge. <br /><br />The family is composed of a father, mother, grandfather, and child. The father is athletic and can cross the bridge in 1 minute; the mother can cross in 2 minutes; the child can cross in 5 minutes; and the grandfather, the slowest, takes 10 minutes to cross. They have a lantern with them. <br /><br />Since it's pitch dark, the bridge can't be crossed without the lantern. The bridge is so narrow that only two can cross at a time, and each pair can only move as quickly as its slowest member. <br /><br />Their pursuer is likely not far behind. What is the quickest way to get everyone across the bridge?<br /><br /><br /><a name='more'></a><br /><br /><br /><br /><br /><br /><br />Solution:<br />They can all get across in 17 minutes. Thanks to Rob Strong for articulating the solution. <br /><br />F&M cross: 2 min<br />F returns: 1 min<br />GF&C cross: 10 min<br />M returns: 2 min<br />F&M cross again: 2 min.Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com4tag:blogger.com,1999:blog-1147784181901427860.post-25245823639838686492010-05-26T08:18:00.000-07:002010-06-02T14:09:20.915-07:00Dots and RowsThis is an old one my dad told me, though I've also heard it as a CarTalk puzzler before:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_KufTlvMetK4/S_06yuywywI/AAAAAAAAAVI/NscZnBeEVEI/s1600/4-dot+row.png" imageanchor="1" style="clear: left; cssfloat: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" gu="true" height="28" src="http://2.bp.blogspot.com/_KufTlvMetK4/S_06yuywywI/AAAAAAAAAVI/NscZnBeEVEI/s200/4-dot+row.png" width="200" /></a></div><br /><br /><br />Above is a straight row of 4 dots: it is a 4-dot row.<br />Next, we have a set of 13 dots that creates 5 different 4-dot rows.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/_KufTlvMetK4/S_06-D4QDII/AAAAAAAAAVQ/ReFgrwMx9DA/s1600/13+dots.png" imageanchor="1" style="clear: left; cssfloat: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" gu="true" height="165" src="http://4.bp.blogspot.com/_KufTlvMetK4/S_06-D4QDII/AAAAAAAAAVQ/ReFgrwMx9DA/s200/13+dots.png" width="200" /></a> </div><br /><br /><br /><br /><br /><br /><br /><br /><br />Now again create 5 different 4-dot rows, but using only 10 dots. Rows must be straight. Oh, and before you tricksters try giving me a single long string of dots, a 5-dot row does not count as 2 4-dot rows. You can use a dot multiple times, but rows can't overlap.<br /><br /><a name='more'></a><br /><br /><br /><br /><br />Solution:<br />It's a five-pointed star, where every vertex is a dot. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com2tag:blogger.com,1999:blog-1147784181901427860.post-64147598910982392362010-05-19T00:00:00.000-07:002010-05-21T14:03:46.766-07:00Cubic Calendar<em>From <a href="http://www.folj.com/puzzles/">http://www.folj.com/puzzles/</a></em><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://www.babyearth.com/images/images_big/10-8155-01.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="200" src="http://www.babyearth.com/images/images_big/10-8155-01.jpg" width="200" wt="true" /></a><a href="http://www.babyearth.com/images/images_big/10-8155-01.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="200" src="http://www.babyearth.com/images/images_big/10-8155-01.jpg" width="200" wt="true" /></a></div>A corporate businessman has two cubes on his office desk. Every day he arranges both cubes so that the front faces show the current day of the month. <br /><br />What numbers are on the faces of the cubes to allow this? <br /><br />Note: You can't represent the day "7" with a single cube with a side that says 7 on it. You have to use both cubes all the time. So the 7th day would be "07".<br /><br /><a name='more'></a><br /><br /><br /><br /><br />Solution thanks to Rob Strong:<br />you need two 1's, and two 2's; you only need one 0, 3, 4, 5, 6, 7, 8, 9. <br /><br />That's twelve digits you need, which is very convenient. <br /><br />However, you need to be able to display nine different days beginning with 0, ten different days beginning with 1, and ten beginning with 2. Because you can't fit ten numerals on a single cube, you'll need to have 0, 1, and 2 represented on both cubes. We knew that about 1 and 2 (since we'll need to display 11 and 22), but now we have a problem, because two 0's makes thirteen numerals needed, and only twelve faces on which to fit them. <br />HOWEVER! We are saved by the rotational symmetry of 6 and 9, and the fact that February 96th only comes around once every ten billion years, far outside the scope of our businessman's lifespan. <br />Thus the dice read as follows:<br /><br />First cube:<br />0<br />1<br />2<br />3<br />4<br />5<br /><br />Second Cube<br />0<br />1<br />2<br />6 (flip to read 9)<br />7<br />8Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com4tag:blogger.com,1999:blog-1147784181901427860.post-70643327538237986892010-05-13T00:00:00.000-07:002010-05-21T13:59:16.893-07:00A Lost and Hungry Vagabond<div class="separator" style="clear: both; text-align: center;"><a href="http://dumbonyc.com/images/blog/baguette061016_450.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="214" src="http://dumbonyc.com/images/blog/baguette061016_450.jpg" width="320" wt="true" /></a></div><br /><br /><em>This is from the </em><a href="http://cartalk.com/content/puzzler/transcripts/200939/index.html"><em>Car Talk Puzzler:</em></a><br />A lost and hungry vagabond happened upon a pair of travelers one of whom had three loaves of bread while the other had five. All of the loaves were the same size and weight. <br /><br />The two travelers decided to share their bread with the vagabond, and that the eight loaves should be shared equally among the three of them. When they had finished, the vagabond reached into his pocket and pulled out eight coins. He handed three coins to the traveler who had had the three loaves and five to the other one and disappeared into the inky shadows. <br /><br />The next morning, right after no breakfast, the one who had received the three coins said to the other one, 'I don't think he should have given three coins to me and five to you. It's not fair.' And he was right. How should the coins have been split up?<br /><br /><a name='more'></a><br /><br /><br /><br /><br />Solution:<br />The travelers sharing their bread should only be compensated for what they gave to the vagabond, rather than what they contributed to the pot - some of that contribution each traveler ate himself. Since the 3 diners shared 8 loaves between them, each one ate 8/3, or 2 2/3 of a loaf. Since the 3-loaf traveler put in 3 loaves but took out 2 2/3 for himself, in effect he only gave 1/3 of a loaf to the vagabond. On the other hand, the 5-loaf traveler put in 5 loaves and took out 2 2/3 for himself, and thus donated 2 1/3 loaves to the vagabond. His 2 1/3 donation was 7 times as big as the other traveler's 1/3 donation, so he deserves 7 coins and the other deserves 1. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com2tag:blogger.com,1999:blog-1147784181901427860.post-2427637626132164382010-04-28T00:00:00.000-07:002010-05-13T07:13:12.175-07:00The Camels<table border="0" cellpadding="0" cellspacing="0"><tbody><tr><td style="text-align: center;"><table border="0" cellpadding="0" cellspacing="0"><tbody><tr><td><img height="33" src="http://www.folj.com/puzzles/media/camels.jpg" width="44" /></td><td><img height="33" src="http://www.folj.com/puzzles/media/camels.jpg" width="44" /></td><td><img height="33" src="http://www.folj.com/puzzles/media/camels.jpg" width="44" /></td><td><img height="33" src="http://www.folj.com/puzzles/media/camels.jpg" width="44" /></td><td width="32"></td><td><img height="33" src="http://www.folj.com/puzzles/media/camels2.jpg" width="44" /></td><td><img height="33" src="http://www.folj.com/puzzles/media/camels2.jpg" width="44" /></td><td><img height="33" src="http://www.folj.com/puzzles/media/camels2.jpg" width="44" /></td><td><img height="33" src="http://www.folj.com/puzzles/media/camels2.jpg" width="44" /></td></tr></tbody></table></td><td style="text-align: center;"><br /></td><td style="text-align: center;"><br /></td><td style="text-align: center;"><br /></td><td style="text-align: center;" width="32"><br /></td><td style="text-align: center;"><br /></td><td style="text-align: center;"><br /></td><td style="text-align: center;"><br /></td><td style="text-align: center;"><br /></td></tr></tbody></table><div class="problem"><br /></div><div class="problem"><i>Borrowed from <a href="http://www.folj.com/puzzles/">http://www.folj.com/puzzles/</a></i> </div><div class="problem"><br /></div><div class="problem">Four tasmanian camels traveling on a very narrow ledge encounter four tasmanian camels coming the other way.</div><div class="problem">Tasmanian camels never go backwards, especially when on a precarious ledge. The camels will climb over each other, but only if there is a camel sized space on the other side.</div><div class="problem">The camels didn't see each other until there was only exactly one camel's width between the two groups.</div><div class="problem">How can all camels pass, allowing both groups to go on their way, without any camel reversing?</div><div class="problem"><br /></div><div class="problem"><i>Hint: to help visualize, use paper clips or coins. </i> <br /><a name='more'></a></div><div class="problem"></div><div class="problem"></div><div class="problem"></div><div class="problem"></div><div class="problem"><br /><br /><br /><br />Solution:</div><div class="problem">Let the camels on the left be 1> 2> 3> and 4>. Let the camels on the right be <A <B <C <D. Arrow indicates direction the camel is facing. One possible solution (although I think the only other solution is symmetrical to this one - let me know if you find others):</div><div class="problem"></div><div class="problem">1> 2> 3> 4> ..... <A <B <C <D</div><div class="problem">1> 2> 3> 4> <A ..... <B <C <D <br />1> 2> 3> ..... <A 4> <B <C <D<br />1> 2> ..... 3> <A 4> <B <C <D<br />1> 2> <A 3> ..... 4> <B <C <D<br />1> 2> <A 3> <B 4> ..... <C <D<br />1> 2> <A 3> <B 4> <C ..... <D<br />1> ..... <A 2> <B 3> <C 4> <D<br />..... 1> <A 2> <B 3> <C 4> <D<br />..... .... <A 1> <B 2> <C 3> <D 4><br />..... .... <A .... <B 1> <C 2> <D 3> 4><br />..... .... <A .... <B .... <C 1> <D 2> 3> 4><br />..... .... <A .... <B .... <C .... <D 1> 2> 3> 4><br />..... <A .... <B .... <C .... <D .... 1> 2> 3> 4><br /><A <B <C <D .... 1> 2> 3> 4><br /></div>Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com5tag:blogger.com,1999:blog-1147784181901427860.post-51388410726430386552010-04-21T07:09:00.000-07:002010-04-21T13:14:24.543-07:00Apples and Oranges<div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_KufTlvMetK4/S88GKakkt4I/AAAAAAAAAUY/IhmBcqlXsDA/s1600/Apples+and+Oranges.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_KufTlvMetK4/S88GKakkt4I/AAAAAAAAAUY/IhmBcqlXsDA/s320/Apples+and+Oranges.png" wt="true" /></a></div><br /><em>This also came from <a href="http://www.ocf.berkeley.edu/~wwu/riddles/easy.shtml">Wu Riddles</a></em>. <br /><br />There are three closed and opaque cardboard boxes. One is labeled "APPLES", another is labeled "ORANGES", and the last is labeled "APPLES AND ORANGES". You know that the labels are currently misarranged, such that no box is correctly labeled. You would like to correctly rearrange these labels. To accomplish this, you may see only one fruit from one of the boxes. Which box do you choose, and how do you then proceed to rearrange the labels?Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com4tag:blogger.com,1999:blog-1147784181901427860.post-37260874741900488852010-04-14T07:04:00.000-07:002010-04-21T06:51:26.249-07:00The Yukiad Contraption<div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_KufTlvMetK4/S8XLJ7r0HgI/AAAAAAAAAT0/ZWix98yBxoQ/s1600/led_hula_hoop_sideways.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_KufTlvMetK4/S8XLJ7r0HgI/AAAAAAAAAT0/ZWix98yBxoQ/s320/led_hula_hoop_sideways.jpg" wt="true" /></a></div><br /><br />Thanks to Jon Campbell for telling me about this one, found at Steven Landsburg's blog, <a href="http://www.thebigquestions.com/2010/04/05/the-yukiad-perpetual-motion-and-me/">The Big Questions</a>. The original puzzle is from the novel <em>The Yukiad</em>, by David Snaith. <br /><br /><br />Consider a glass contraptionâ€”a perpetual motion machine, reallyâ€”consisting of a clear glass hula hoop on the ground containing several colored beads, which travel through the hoop, some clockwise, some counterclockwise, all at the same speed, bouncing off each other in perfectly elastic collisions whenever they collide. Whenever two beads collide, they instantly bounce off each other and proceed in the opposite of their original directions, still at the same speed. Take a snapshot of this system at, say, 12PM. Must there be some time in the future when another snapshot of the system will look identical? In other words, does the history of the system repeat itself?<br /><br /><em>(Hint: for simplicity, start with a few beads, then generalize)</em><br /><br /><a name='more'></a><br /><br /><br /><br /><br />Solution (thanks again to Jon):<br /><br />Instead of being beads, you can pretend that the system consists of men with different colored hats walking around in a circle at a constant speed, where each man turns around each time he encounters another man. Then the question would be, would there ever be a time when all men are at the same place where they started? But that question is equivalent to the following: instead of turning around each time a man comes across another man, he will walk right by him, but trade hats with him. From the perspective of someone looking at these men from above, this "walk right by but trade hats" scenario (call it A) will look identical to the "turn around and don't trade" scenario (call it B), and the question is, will there ever be a time when all the hats are in the same place as at the beginning.<br />Using scenario A, it is obvious that after a particular man has walked the entire circle, all of the other men have done the same, and are in their starting positions. But, each may not be wearing the hat he started with, so the proof is not yet done. But even if they aren't wearing the same hats as they were in the beginning after 1 rounding of the circle, they will after some integer number of roundings n. This seems intuitive, and the best way I can explain it is with an example: suppose a red hat started with person X, and after 1 rounding, it is with person Y. It is clear that after some # or roundings, the red hat will end up back with person X -- i.e. it will not end up looping between some group of people that doesn't include X, since there would be no way for it to enter that loop in the first place if it existed. Suppose that for this red hat, the required minimum # of roundings before getting back to person X is 4. Suppose that for some other hat, the minimum # of roundings is 7, and so on, different #s for each hat (but actually there will be various hats that share the same minimum # of roundings). Then, at the very latest, the minimum # of roundings n for which ALL hats are back with their original person is 4*7*(...for all hats). So we know that there exists a # n of roundings after which all hats are back where they started.Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.comtag:blogger.com,1999:blog-1147784181901427860.post-66724523043597376932010-04-07T12:17:00.000-07:002010-04-14T06:38:04.660-07:00Cut the Cake<div class="separator" style="clear: both; text-align: center;"><a href="http://www.cakecrafts.co.uk/cdata/26961/img/26961_1550266.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="240" nt="true" src="http://www.cakecrafts.co.uk/cdata/26961/img/26961_1550266.jpg" width="320" /></a></div><br />Happy Birthday, Kyle! <br />With 3 straight cuts through a cylindrical cake, make 8 equal-sized slices. <br /><br />There are two solutions.<br /><br />(<em>Hint: you are allowed to move some or all of the cake)</em><br /><br /><a name='more'></a><br /><br /><br /><br /><br />Solution:<br />Both solutions involve making two normal vertical cuts to make quarters. For the third cut, one way is to hold the knife parallel to the ground and cut the cake into top and bottom halves. However this results in some people not getting any frosting. The alternative is to rearrange the quarters - either stack them on top of each other, or line them up in one long line - so that you can cut through all of them with one more long vertical cut. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com7tag:blogger.com,1999:blog-1147784181901427860.post-27230461144265067602010-03-31T00:00:00.000-07:002010-04-07T12:06:35.960-07:00The Prisoner, the Liar, and the Truth, Part II: Enter the Other Guy<div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/_KufTlvMetK4/S7KHsgWNlzI/AAAAAAAAATU/2z-GeLmCmQU/s1600/Prison+Doors.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" nt="true" src="http://4.bp.blogspot.com/_KufTlvMetK4/S7KHsgWNlzI/AAAAAAAAATU/2z-GeLmCmQU/s320/Prison+Doors.jpg" /></a></div>Recall from a couple weeks ago, the door that leads to the electric chair in the Prison for Creative and Unusual Punishment. It is located in a depressing, windowless basement hallway near the back, right next to an identical door that leads to an unguarded fire exit. <br /><br />A prisoner with similar lamentable circumstances to the guy in part 1 was being brought in for execution, and the warden wanted to give him one last chance at freedom. This time, however, would be more complicated. Rather than two guards, there were three guards with them, named Al, Bob, and Carl. The warden held a brief huddle with the three guards, out of earshot of the prisoner. He then told the prisoner some of what went down in the huddle. <br /><br />"I've instructed one guard to tell nothing but lies when asked a yes-or-no question. I've instructed another guard to tell only the truth when asked a yes-or-no question. I then repeated one of those two instructions to the remaining guard. Unfortunately, I'm not going to tell you whether there are two liars an a truth-teller, or two truth-tellers and a liar. And I'm certainly not going to tell you which is which. You have two yes-or-no questions to ask, choosing one guard at a time to respond (no asking a question of the whole crowd). From the information you glean, you may choose a door. Best of luck."<br /><br />What two yes-or-no questions can the prisoner ask whose answers will lead him to freedom?<br /><br />Please submit answers in the comments section of the blog, or to me directly (no spoilers in Google Buzz).<br /><br /><a name='more'></a><br /><br /><br /><br /><br /><strong>Solution:</strong><br />The wording here is kind of tricky, so bear with me. <br /><br /><strong>Question 1:</strong><br />It doesn't matter whom you ask the first question, so we will arbitrarily pick Al and ask him the following: "Would Bob say that Carl tells the truth?"<br /><br />You can use the answer to this question to determine whether you have two liars or two truth-tellers. No matter whom you ask this type of question, or which guard is which, the meaning of the answers will be the same. If there are two liars, the answer will be 'yes'. If there is one liar, the answer will be 'no'. Try any combination or orientation of liars and truth tellers, this will be true. <br /><br /><strong>Question 2:</strong><br />Again it doesn't matter whom you ask, but for consistency we will pick Al again and ask him, "Would Bob say that Carl would say the door on the left leads to freedom?" If there are two liars, they will cancel each other out and the answer will be true. In this case, you should go through the door on the left. If there is only one liar, the answer will be false, and you should go through the door on the right. <br /><br /><strong>Explanation</strong><br />By asking the second question through all three guards, you get an answer that you can know is true or false regardless of the order of the guards being asked. A lie about a truth and a truth about a lie both end up being false statements. This is the principle that led the prisoner to freedom in part 1 of this riddle - the prisoner asked guard A what guard B would say about which door leads to freedom. No matter which guard was the liar - A or B - the answer still came out a lie. <br /><br />This principle can be equated to multiplying positive and negative numbers. Just like lie about a true statement is false, a negative (-) times a positive (+) is negative (-). This property is commutative - the order doesn't matter: (+)*(-) = (-)*(+) = (-). Furthermore, a lie about a lie is a truth, just like a negative times a negative is a positive. <br /><br />So when you ask the second question through all three guards, the answer is true if there are two liars, as modeled by the equation (-)*(-)*(+) = (+). Or the answer is false if there is only one liar: (-)*(+)*(+)=(-). This is true regardless of order. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com3tag:blogger.com,1999:blog-1147784181901427860.post-65770065887284841292010-03-29T11:19:00.000-07:002010-03-31T06:43:33.505-07:00The Catenary Chain<div class="separator" style="clear: both; text-align: center;"><a href="http://upload.wikimedia.org/wikipedia/commons/0/04/Kette_Kettenkurve_Catenary_2008_PD.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="240" nt="true" src="http://upload.wikimedia.org/wikipedia/commons/0/04/Kette_Kettenkurve_Catenary_2008_PD.JPG" width="320" /></a></div><br /><em><span id="goog_1246791214"></span><a href="http://opinionator.blogs.nytimes.com/2010/03/28/power-tools/?hp">This week's post<span id="goog_1246791215"></span> on Steve Strogatz's New York Times column</a> reminded me of this puzzle. In it he re-explains math in a creative and intuitive way, from basic counting through imaginary numbers, functions, and more. The post reminded me of this because in it he talks about how mathematical functions can explain every-day shapes, such as how water at a drinking fountain forms a parabola. A hanging chain forms a <a href="http://en.wikipedia.org/wiki/Catenary">catenary</a>, but its shape can be approximated with a parabola. It is a great new one I found at </em><a href="http://www.ocf.berkeley.edu/~wwu/riddles/easy.shtml"><em>Wu Riddles</em></a><em>, and apparently it originated at a Microsoft Interview.</em> <em>BUT: don't be intimidated by all the math, I have faith that you can solve this one. </em><br /><br />You have a 6-foot long chain that is suspended at its ends, tacked to a wall. The tacks are parallel to the floor. Due to gravity, the middle part of the chain hangs down below the ends, forming a 'U'-type shape; the height of this 'U' is 3 feet from top to bottom. Find the distance in between the tacks.<br /><br /><a name='more'></a><br /><br /><br /><br /><br />Solution:<br />As the commenters pointed out below, the distance between the tacks is zero. The only way for the dip in the chain to be three feet is for the chain to drop 3ft straight down, then 3ft straight up. There is no chain left to spare for horizontal distance. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com5tag:blogger.com,1999:blog-1147784181901427860.post-79613008007864108232010-03-17T07:01:00.000-07:002010-03-25T13:24:04.289-07:00The Prisoner, the Liar, and the Truth<div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_KufTlvMetK4/S6DgfP0GtRI/AAAAAAAAASk/9hLu88N1v9c/s1600-h/Prison+Doors.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/_KufTlvMetK4/S6DgfP0GtRI/AAAAAAAAASk/9hLu88N1v9c/s320/Prison+Doors.jpg" vt="true" /></a></div>The door that leads to the electric chair in the Prison for Creative and Unusual Punishment is located in a depressing, windowless basement hallway near the back. It just so happens that it is right next to an identical door that leads to an unguarded fire exit. <br /><br />One day the warden was bringing a prisoner to be executed, but was feeling sorry for him. The prisoner had been convicted on dubious charges and DNA evidence had been unearthed that might have exonerated him, but the court threw it out. So the warden wanted to give the prisoner one more chance at freedom. There were two guards with them and the warden whispered in their ears, out of earshot of the prisoner. The warden instructed one to tell nothing but lies when asked a yes-or-no question and the other to tell only the truth when asked a yes-or-no question. Otherwise they were to remain silent. The prisoner couldn't tell which was the liar and which was the truth-teller. <br /><br />The warden then told the prisoner that he could ask <u>one</u> yes-or-no question to <u>one</u> of the guards. If, from the answer, he could pick the correct door to freedom, the guards would look the other way. <br /><br />What question should the prisoner ask?<br /><br /><a name='more'></a><br /><br /><br /><br /><br />Solution:<br />Point to a door and say to one guard, "If I asked the <em>other</em> guard if this were the door to freedom, would he say yes?" The answer will be a lie no matter whom you ask. Even the truth-teller, when speaking on behalf of the liar, would be forced to give a false answer. So if the answer is yes, choose the other door. If the answer is no, choose the door you pointed to. Charlie Guthriehttp://www.blogger.com/profile/15624106792764866509noreply@blogger.com2