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Your Favorite Math/Logic Riddles? 1965

Posted by Cliff
from the top-10-mind-twisters dept.
shma asks: "Whether you're involved in the Sciences, Mathematics, or Engineering, you undoubtedly enjoy finding simple solutions to seemingly difficult problems. I'm sure you all have a favorite mind-bender, and who better to share it with than the Slashdot community? Post your own problems and try to solve others. Just one request: If you have figured out the solution, link to it in a post, rather than write it out where anyone can see it." What brain benders tickle your fancy?
"Here's a sample to consider: You're in a dark room with 50 quarters, 18 of which are heads up. You are allowed to move around the coins or flip some or all of them, if you wish. Problem is, it's too dark to tell what you're moving or flipping (no, you can't figure it out by touch either). Your job is to split the coins into two groups, each of which has the same number of heads up coins. How do you accomplish this?"
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Your Favorite Math/Logic Riddles?

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  • Soduku (Score:4, Interesting)

    by beacher (82033) on Sunday October 16, 2005 @12:37AM (#13800847) Homepage
    They drive me nuts. Array and vecor logic. Fun

    -B
  • by Paladine97 (467512) on Sunday October 16, 2005 @12:38AM (#13800854) Homepage
    I wouldn't say I have a favorite problem but often when I'm bored I'll pen down the Pythagorean theorem and solve it manually. 0 = ax*x + bx + c. I'll work it out until I get the solution that (I hope) everybody knows and loves! It helps to keep my math skills alive during boring meetings.
  • by Omnieiunium (872399) <canadiancanuck@@@gmail...com> on Sunday October 16, 2005 @12:38AM (#13800857) Journal
    Is obviously 42
  • easy one (Score:5, Funny)

    by zanderredux (564003) on Sunday October 16, 2005 @12:39AM (#13800859)
    prove that a^n=b^n+c^n for any n.
  • Sequence (Score:2, Interesting)

    by blystovski (525004)
    What is the next line in the following sequence? 1 11 21 1211 111221 312211
  • by heinousjay (683506) on Sunday October 16, 2005 @12:40AM (#13800866) Journal
    Turn a light on.
    • by radtea (464814) on Sunday October 16, 2005 @10:50AM (#13803240)
      Turn a light on.

      I was once a judge at a "Phyics Olympics" where there was one puzzle in which students had to figure out the wiring if a circuit consisting of a couple of light bulbs and a couple of switches. They were "supposed" to solve the puzzle by flipping the switches, noting what lights were on and off, and inferring the circuit.

      One team took the apparatus apart and inspected the wiring.

      I gave 'em full marks.

      The head judge went spare.

      Science is not a game, and there aren't any rules according to which you are "supposed" to solve the problem. Alexander the Great was demonstrating the practice of experimental science when he unravelled the Gordian knot, and Feyrabend was onto something when he said, "Anything goes."

      Puzzles set by humans have more to do with communication between the puzzle-setter and the puzzle-solver than anything else. Some people even decry computer-generated puzzles because of this--they say that the pleasure they get from solving puzzles comes from the feeling of interaction with another mind.
  • Petals of the Rose (Score:5, Interesting)

    by Alien54 (180860) on Sunday October 16, 2005 @12:41AM (#13800871) Journal
    I personally like the petals of the rose [google.com]

    Bill Gates is said to have solved the problem by memorizing the combinations first [borrett.id.au], the brute force approach.

    It ones of those that requires a knack for seeing the simple things

    • Ah I got it. Took 16 rolls (written down) and almost hour. The fact that I am ignorant about roses didn't help :)

      Anyway I read that story and it didn't appear to me that he was trying to solve it by memorization, but rather that after an hour, seeing hundreds of rolls, he remembered many of them, which isn't all that surprising. What I got out of the story, is that he persistently kept at the problem trying many different ideas until he finally got it, even after everyone else in the group had solved it or
  • What is the degree of the angle between the hour hand and the minute hand when it is 2:15?

    If you have a piece of paper, and you draw any quadrilateral of any size (rhombus, rectangle, or square) on that piece of paper. How can u divide that piece of paper in half so that it also evenly divides teh quadrilateral?

    If you have a 5 gallon jug and a 3 gallon jug of water, and a hose so u can refill any as u please. What are the steps to get exactly 4 gallons of water?

    • Jugs (Score:3, Interesting)

      by Noksagt (69097)
      If you have a 5 gallon jug and a 3 gallon jug of water, and a hose so u can refill any as u please. What are the steps to get exactly 4 gallons of water?
      Fill the 5G jug. Pour it into the 3G jug, so you have 2G in the 5G jug. Empty the 3G jug & pour the 2G from the 5G jug into the 3G jug. Refill the 5G jug & finish filling the 3G jug. It will only take 1G, so you will have 4G in the 5G jug.
    • Jugs-

      Fill the 3G and pour it into the empty 5G. Refill the 3G and pour into the 5G until full, this leaves 1 gallon in the 3G. Empty the 5G and pour the contents of the 3G (1 gallon) into the 5G. Now fill the 3G and pour completely into the 5G, making 4 gallons total in the 5G.
    • angle answer (Score:3, Informative)

      by mikeage (119105)
      What is the degree of the angle between the hour hand and the minute hand when it is 2:15?
      22.5 degrees.

      Yes, you can do it iteratively until inifinity, but the minute hand is at 90 degrees off 12, and the hour hand is at 60 for 2, plus 30/4 for the :14, = 67.5. The difference is 22.5
  • by Doc Ruby (173196) on Sunday October 16, 2005 @12:47AM (#13800903) Homepage Journal
    42
  • I haven't thought about a more efficient way, but it seems that the surest bet would be to split the coins into two equal groups, and then flip all the coins in each group. You're most likely to end up with 25 heads up coins in each group (being the maximum entropy state).
    • by Stalus (646102) on Sunday October 16, 2005 @01:16AM (#13801094)

      Simply place any 18 coins into the second group and flip those over.

      If you flip a coin over that was heads, it is now tails and is eliminated from consideration. If you flip a coin over that was tails, it marks with heads a coin selected that was not heads. Therefore after 18 coins are flipped, the number of heads in the second pile is equal to the number of heads that are left in the first pile.

  • by bratwiz (635601)
    Here is the little brain teaser I thought up-- see if you can solve
    it...

    In the following sequence:

    1, 4, 8, 13, 21, 30, 36, 44...

    What is the next number and why:

    A. 48

    B. 50

    C. 53

    D. 57

    E. 61

    F. There is no pattern
  • f(x)=x^(1/x)

    It is not defined for negative values or 0. It is defined only for x>0. At 1, f(x)=1, then it peaks somwehere in 2.71
    What I'm really interested in is the first derivative. Where f'(x) is 0, is the maximum of f(x). Just one catch: no limits in the formula. I don't want something that I need to calculate forever; I want a formula giving me a value I can calculate to an arbitrary precision.

    The problem comes into play with the (1/x) in the exponent. All attempts to derive this result in a
  • Just to get an easy, universally-agreed-upon one out of the way.... .9999 =! 1

    Now we can move on to questions that can generate some real debate.
  • by LeonGeeste (917243) * on Sunday October 16, 2005 @12:57AM (#13800964) Journal
    I met a man with seven wives. Every wife had seven sacks, and every sack had seven cats, and ever cat had seven kits. Kits, cats, sacks, and wives, how many were going to St. Ives?
  • by Quirk (36086) on Sunday October 16, 2005 @12:58AM (#13800976) Homepage Journal
    Since I tend to muck about in philosophy, history and epistemology I'll go with perhaps the most ancient riddle.

    Epimenides was a Cretan who made one immortal statement: "All Cretans are liars."

    "The Epimenides paradox [wikipedia.org] is a problem in logic. This problem is named after the Cretan philosopher Epimenides of Knossos (flourished circa 600 BC), who stated , "Cretans, always liars". There is no single statement of the problem; a typical variation is given in the book Gödel, Escher, Bach (page 17), by Douglas R. Hofstadter.

  • Truth vs. Lies (Score:4, Insightful)

    by sheetsda (230887) <doug.sheets@gmai l . c om> on Sunday October 16, 2005 @01:01AM (#13800987)
    You find yourself before indistinguishable two doors, each with a statue. One door will lead to salvation, the other to death. The statue that guards the door to salvation always tells the truth, the statue to the door to death always lies. You may pose only one question to only one statue. What do you ask to determine which door is which?

    Answer(ROT13): Nfx nal dhrfgvba gb juvpu lbh nyernql xabj gur nafjre. Gb qrgrezvar juvpu qbbe vf juvpu lbh arrq gb xabj gur eryngvbafuvc bs gur nafjre lbh ner tvira gb gur gehgu. Gur guvat V yvxr nobhg guvf evqqyr vf vg sbeprf lbh gb pbafvqre gur bcrengbe va gur ybtvpny fgngrzrag gb or gur inevnoyr. Nqqvgvbanyyl crbcyr nera'g hfrq gb nfxvat dhrfgvbaf jura gurl nyernql xabj gur nafjre fb gurl graq abg gb or noyr gb guvax bs n fbyhgvba evtug njnl. Gur jubyr guvat orpbzrf boivbhf jura lbh cbfr n dhrfgvba fhpu nf "Ner gurer gjb fgnghrf urer?"
  • First up, does Goedel's incompleteness theorem imply that computers will never be able to have human-like intelligence [wikipedia.org]?

    Other things I like are not necessarily problems, but things that just inspire awe, such as proving that .99... = 1, or that formula that shows pi*log^-1 = 0 or whatever it is... is always struck me as the Grand Unification Theory of algebra and geometry. It's so simple and shows that these numbers, which are so hard for me to work with, combine in some fashion to show some property that i
  • Lightbulb problem (Score:5, Interesting)

    by Ellen Spertus (31819) on Sunday October 16, 2005 @01:02AM (#13800996) Homepage
    Given:
    • One room has three switches, labeled A, B, and C.
    • Another room has three light bulbs, labeled 1, 2, and 3.
    • Each switch is connected to one bulb, but you do not know which is connected to which.
    • When inside either room, you cannot see the other room.
    • You begin in the room with the switches and may turn the switches on and off in any way you choose.
    • Once you leave the room with the switches, you may not reenter it. You may, however, go to the room with the light bulbs.
    How can you determine which switch is connected to which light? Here is a hint [nicemice.net] and solution [nicemice.net].

    I like this problem because people are ordinarily good at logic have so much trouble with it. I once had the pleasure of meeting Donald Knuth and stumped him with this puzzle.

    • by TheVoice900 (467327) <kamil&kamilkisiel,net> on Sunday October 16, 2005 @05:45AM (#13802212) Homepage
      I know lots of people have commented on using the hot/cold method to determine which bulb is which, there's another problem with that as well: You don't know the initial state of the bulbs.

      Say for example all the bulbs are initially ON, and you flip two of the switches to what you think is on. Then when you flip one of them to what you think is "off" and wait a while, and go in to the room, you'll find two bulbs on, but you'll misidentify them because the one you thought you switched to "off" you actually turned "on". Not to mention they could be in mixed states initially..
    • Since I have LED lighting in my house, this won't work! And if the world switched to solid state lighting, this puzzle will be obsolete.

      BTW: Donald Knuth probably hates you.

  • Sticky Triangles (Score:5, Interesting)

    by Doc Ruby (173196) on Sunday October 16, 2005 @01:15AM (#13801090) Homepage Journal
    Let's say I have a stack of sticks: all identical, inflexible, unbreakable. Sticks can touch only at their ends, not in between.

    If I give you 3 sticks, you can make one triangle. If I give you 2 more sticks (5), you can make 2 triangles. If I give you another stick (6), how can you make 4 triangles?
  • by nyri (132206) on Sunday October 16, 2005 @01:16AM (#13801098)
    Is ..:: Riddles ::.. [berkeley.edu]. In has (amogst others) the famous "prison with a lamp" problem:

    100 prisoners are imprisoned in solitary cells. Each cell is windowless and soundproof. There's a central living room with one light bulb; the bulb is initially off. No prisoner can see the light bulb from his or her own cell. Each day, the warden picks a prisoner equally at random, and that prisoner visits the central living room; at the end of the day the prisoner is returned to his cell. While in the living room, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting the claim that all 100 prisoners have been to the living room. If this assertion is false (that is, some prisoners still haven't been to the living room), all 100 prisoners will be shot for their stupidity. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world can always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity.

    Before this whole procedure begins, the prisoners are allowed to get together in the courtyard to discuss a plan. What is the optimal plan they can agree on, so that eventually, someone will make a correct assertion?
  • by brian0918 (638904) <brian0918NO@SPAMgmail.com> on Sunday October 16, 2005 @01:20AM (#13801120)
    There is a king and there are his n prisoners. The king has a dungeon in his castle that is shaped like a circle, and has n cell doors around the perimeter, each leading to a separate, utterly sound proof room. When within the cells, the prisoners have absolutely no means of communicating with each other.

    The king sits in his central room and the n prisoners are all locked in their sound proof cells. In the king's central chamber is a table with a single chalice sitting atop it. Now, the king opens up a door to one of the prisoners' rooms and lets him into the room, but always only one prisoner at a time! So he lets in just one of the prisoners, any one he chooses, and then asks him a question, "Since I first locked you and the other prisoners into your rooms, have all of you been in this room yet?" The prisoner only has two possible answers. "Yes," or, "I'm not sure." If any prisoner answers "yes" but is wrong, they all will be beheaded. If a prisoner answers "yes," however, and is correct, all prisoners are granted full pardons and freed. After being asked that question and answering, the prisoner is then given an opportunity to turn the chalice upside down or right side up. If when he enters the room it is right side up, he can choose to leave it right side up or to turn it upside down, it's his choice. The same thing goes for if it is upside down when he enters the room. He can either choose to turn it upright or to leave it upside down. After the prisoner manipulates the chalice (or not, by his choice), he is sent back to his own cell and securely locked in.

    The king will call the prisoners in any order he pleases, and he can call and recall each prisoner as many times as he wants, as many times in a row as he wants. The only rule the king has to obey is that eventually he has to call every prisoner in an arbitrary number of times. So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don't know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose.

    Here's one last monkey wrench to toss in the gears, though. The king is allowed to manipulate the cup himself, k times, out of the view of any of the prisoners. That means the king may turn an upright cup upside down or vice versa up to k times, as he chooses, without the prisoners knowing about it. This does not mean the king must manipulate the cup any number of times at all, only that he may.

    Assume that both the king and the prisoners have a complete understanding of the game as I have just explained it to you, and that the prisoners were given time beforehand to come up with a strategy. The king was able to hear the prisoners discuss, however, so also assume that if there is a way to foil a strategy, the king will know it and exploit the weakness. The prisoners must utilize a strategy that works in absolutely every single possible case.

    Now you must figure out not only how to keep the prisoners alive, but how to also ensure their eventual freedom. When can any one of them be certain they've all been in the central chamber of the dungeon at least once? And how? Don't try to imagine any trickery like scratching messages in the soft gold of the chalice. The problem is as simple as it sounds. The prisoners have absolutely no way of communicating with each other except through the two orientations of the chalice. If any of them attempts any trickery at all they will all be beheaded. All the prisoners can do is turn the chalice upside down or right side up, as they choose, whenever they are called into the chamber.



    (written by a former roomate)
    • It seems that the Chalice is a red-herring.

      Obviously, the most devious thing that the king can do is to always make sure the chalice is rightside up. Therefore, it will always be rightside up. Therefore, the chalice can provide no information. Therefore, it is a red-herring.

      With all other avenues of gathering information forbidden, there seems no information left to base an answer on.
    • MOD PARENT TROLL (Score:4, Informative)

      by LeonGeeste (917243) * on Sunday October 16, 2005 @04:24PM (#13805159) Journal
      I had a conversation with Brian0918 on AIM this morning, in which he revealed he's really trolling when I pointed out to him there is no solution (see my other posts) on this topic. Here's a little tidbit: ". i usually just post the problem to get people into big disputes, which so far has worked 2 out of 2 times". If you want the full conversation, email me at sbartaNOSPAM_at_MAPSONgmail.com.
  • hats (Score:3, Interesting)

    by blackcoot (124938) on Sunday October 16, 2005 @01:20AM (#13801121)
    let's see... the problem goes roughly like this:

    you have five hats (two red, three black) and three people. you queue the people up in order of height and have them face the same way (this way the tallest person can see the two people in front of him/her, the middle person can see the shortest person, and the shortest person can't see anyone). you put a hat on each person's head and instruct them that they are not allowed to take the hat off or turn around. you then ask them to tell you what color their hat is. after a while, the person at the front of the line correctly announces the color of his/her hat. how did the person at the front of the line know and what were the other hat colors?
    • Re:hats (Score:3, Interesting)

      by drawfour (791912)
      If the tallest person saw two red hats in front of him, he would immediately know his had was black and say "My hat is black!" He didn't, therefore he either sees two black hats or one black hat and one red hat.

      Knowing this, the second person looks at the hat in front of him. If he sees a red hat, he knows his is black. Since he does not see a red hat, his is either red or black, he doesn't know. But knowing this, the first guy can deduce that his hat is black.

      I don't think you can know what the oth
  • you're given a globe (Score:3, Interesting)

    by rmm4pi8 (680224) <rmiller@reaCOWso ... t minus herbivor> on Sunday October 16, 2005 @01:40AM (#13801254) Homepage
    Assuming the earth is a perfect sphere, describe the solution set of points where you can go 1 mi south, 1 mile east, and 1 mile north and return to your starting point. Hint: the cardinality of the set is R cross Z + 1 (and yes, I know that's equal to R, but expanding it makes it a more effective hint). Feel free to email me for more hints.
  • by Council (514577) <`moc.liamg' `ta' `eornumr'> on Sunday October 16, 2005 @11:25AM (#13803399) Homepage
    Oh, woe is me. I have a perfect logic puzzle, but was unlucky enough to be otherwise engaged when this story was posted. (By the way: a soft couch, a carefully selected DVD, half a bottle of rum, and a girl. Guess which element to this excellent scenario was fucking ruined by copy protection? I'll give you a hint: I may have just switched sides in this movie piracy debate. Fuck the RIAA. It was a perfectly legal store-bought DVD. Fuck them all.)

    But anyway, logic puzzles. This logic puzzle is excellent. I've had it up on my site (http://www.xkcd.com/blue_eyes.html [xkcd.com]), and after I got boingboing'ed I got a lot of email about it, so I've been able to tweak the wording to get rid of most of the confusing stuff, leaving only the logic. It's extremely subtle; I've never seen anything like it.

    Here's the puzzle:

    A group of people live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. If anyone has figured out the color of their own eyes, they [must] leave the island that midnight.

    On this island live 100 blue-eyed people, 100 brown-eyed people, and the Guru. The Guru has green eyes, and does not know her own eye color either. Everyone on the island knows the rules and is constantly aware of everyone else's eye color, and keeps a constant count of the total number of each (excluding themselves). However, they cannot otherwise communicate. So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes, but that does not tell them their own eye color; it could be 101 brown and 99 blue. Or 100 brown, 99 blue, and the one could have red eyes.

    The Guru speaks only once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

    "I can see someone with blue eyes."

    Who leaves the island, and on what night?

    There are no mirrors or reflecting surfaces, nothing dumb, It is not a trick question, and the answer is logical. It doesn't depend on tricky wording, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."

    And lastly, the answer is not "no one leaves."
    • *spoiler warning*

      I think this stipulation is also necessary:

      1) That everyone with blue eyes (at least) is wholly involved in figuring out if they have blue eyes and should comply (bear with me, this is different than you think)

      Without this specification, there can be no implicit communication as to the understanding of others.

      But to be fair, this is hardly the end of the specifications, and is why I so detest logic puzzles. An earlier poster had it right when they said that a logic puzzle is hardly about
    • One small correction:
      "The Guru speaks only once (let's say at noon), on one day in all their endless years on the island."
      should read:
      "The Guru speaks only once (let's say at noon), on each day of all their endless years on the island."

      "only once... on one day" says that after the first day the Guru never speaks again.
  • Coloured stamps (Score:3, Informative)

    by slavemowgli (585321) on Sunday October 16, 2005 @12:57PM (#13803962) Homepage
    Here's another nice one, courtesy of Raymond Smullyan.

    Suppose there are three people, called A, B and C. Each of these is a "perfect logician"; that is, given some information, they all are able to immediately draw any and all conclusions that can possibly be drawn from this information. Furthermore, suppose there are four red and four green stamps.

    Now, all three of them close their eyes, and two stamps are glued to their foreheads, each; the remaining two stamps are put away. Now, they all open their eyes again.

    Then, the first, A, is asked whether he knows the colours of the stamps on his forehead. He says he doesn't. Then B is asked the same thing, and also says he doesn't, and afterwards, C is asked and says he doesn't, too. Now, A is asked a second time, and he still says he doesn't know. But then, when B is asked a second time, he now says he does know.

    The question is: how?
  • by BoRegardless (721219) on Sunday October 16, 2005 @01:42PM (#13804247)
    If you really get one of 'those' meetings or classes, you can try this. It is so boring, you have already made another Tic Tac Toe crossed set of lines. Take all 10 numeral digits and put them in the Tic Tac Toe so that all horizontal, all vertical and all diagonal sums each add up to ... 15 I give no hints.

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