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Everything posted by bonanova

  1. Random triangles in a circle

    Kudos to all. @flamebirde set up an attack framework, @Izzy got the puzzle to its knees, @Molly Mae knocked it out for the count and finally @plasmid, who just might have a penchant for shooting rabbits with elephant guns, put a bullet through its head. To sort out the credits I'll don my judge's robes and presume to declare a verdict: @flamebirde told us it's all about angles. (Honorable mention. +1) @Molly Mae made four (equal) quadrants, which I have in my solution, and gave the right answer but didn't prove it, then switched to the @flamebirde - @Izzy model and backed up the answer with a word proof. (FTW) What must have been only moments later, @Izzy put numbers on things but, probably exhausted from her class, got the wrong number. (Honorable mention. +1) @plasmid dazzled us with calculus symbols and cold-blooded math to lay the newly-deceased puzzle to rest. (also Honorable mention, +1 but it doesn't seem enough. Timing is everything.) As I followed your discussion, (credit to you all) it occurred to me that this might be the simplest non-math approach:
  2. Random triangles in a circle

    @Molly Mae and @Izzy you both
  3. Ants on a checkerboard

    Right so far, and the one not yet calculated is interesting.
  4. Green and Yellow hats

    This approach saves |n/2| prisoners, and that is optimal.
  5. On a table are three plates, containing a, b and c jelly beans, in some order, where a < b < c. At any time you may double the number of jelly beans on a plate, by transferring beans to it from a fuller (or equally populated) plate. After one such move, for example, the plates could have 2a, b-a, and c beans. Using a series of these moves, Is it possible to remove all the jelly beans from one of the plates?
  6. Jelly beans join the clean plate club

    That's it. Binary representation of B/A indicates moves to make B smaller than A was. Rinse and repeat. Nice.
  7. Green and Yellow hats

    OK so I see MM quoting an Izzy post, but I don't see that original post. Can I invite Izzy to repost it, with complete description? Because ... it's the solution. Nice solve.
  8. Buried Proverb

    The word buried here has but one letter. Did you find a jelly roll in Gearhardt's Bakery? It's the best one I've ever seen. The rug at her stairway was made in India. He's an old friend. Amos sold his bicycle to an old friend.
  9. Green and Yellow hats

    Red herring (which I think you've figured out)
  10. Random triangles in a circle

  11. Buried Proverb

    Not sure if ... answer is in the title Sell to? Yes. I fixed the OP. Good catch. When I see ... right track.
  12. Place two dimes and two pennies in a line with a space between them like this: P P D D _ _ _ _ _ Using a sequence of moves, switch the two groups of coins to achieve this position: D D P P _ _ _ _ _ Moves are of two types: slide a coin to an (adjacent) empty space. jump a coin over another coin into an empty space. Type (1) move: P P D D _ _ _ _ _ Type (2) move: P D P D _ _ _ _ _ The underlines show the (only) five legal locations for the coins to occupy. What is the smallest number of moves needed?
  13. Green and Yellow hats

    Yes and yes.
  14. Green and Yellow hats

    Hmmm. OP does not rule out movement. But it does rule out communicating. So let's say that if the prisoners want to be at some preferred location in the room, that's permissible. But their chosen location can't be in any way influenced by hat color -- i.e., all movement must occur before the hats are placed.
  15. In a long hallway, 100 prisoners are given red or blue hats, whose color only the other prisoners can see. At a signal given by the warden the prisoners must walk single file through a door and take their places inside a large room. The room is circular and its wall, ceiling and floor are featureless. Nothing is said, nor are any gestures made to prisoners as they enter the room and take their place. When the last prisoner has taken his place the warden inspects the configuration of their hat colors. If the colors form two monotonic groups separable, say, by some straight line, then all the prisoners are freed. If their hat colors instead are intermingled, they are all executed. Prisoners are allowed to discuss strategy before receiving their hats. What is their fate? Let's see, what else? Oh ya, they can't just pass their hats around. They're super-glued on their heads. Ouch. And no one has a magic marker to ... uh ... you know, make a line ... or anything like that.
  16. Green and Yellow hats

    Nope. They act on what they see. Sorry.
  17. Whodunit?

    The Threedie brothers, Al, Bert, Chuck, Dick and Eddie, lived in a cabin 3 miles up the old mountain trail, and it was known they didn't get along all that well. This morning, Eddie was found dead behind the cabin, and his brothers, the only suspects in the case, were being questioned by Inspector Sherlock. It was known that, of the four, at least 3 were absolute truth-tellers, and none of them ever lied and told the truth in a single day. All four, of course, denied murdering their brother. The Inspector started by asking each brother what he had done that morning: Al: I was analyzing random groups of 3 numbers, and I found that if the numbers sum to zero then their product is the average of their cubes. Bert: I was analyzing random polygons with 3 sides, and I found that if I trisected all their angles I could make an equilateral triangle. Chuck: I planted a dozen apple trees out in the orchard, and I found a way to make eighteen rows of 3 trees, each row being dead-on straight. Dick: I went out and ran 3 miles in the woods, and I've figured out that one of my 3 (living) brothers is lying. The Inspector called in these clues to one of his friends at BrainDen, and in 3 shakes of a lamb's tail the case was solved. The sound you hear is your phone ringing. It's your chance to be famous!
  18. Building cars

    So when OP says "build exactly one car a day (no more, no less)" it means you can build any number of cars in the interval [1 2) because "and to be clear a partial car is as good as not building a car." So if you built 1 1/3 cars on the first day it would count as "exactly 1 car," because it would pass 1/3 of a car to the second day, when you would then have to build any number of cars in the interval [2/3, 1 2/3)? In general, is it correct to believe that at the end of every nth day you must have built [n, n+1) cars, except for n=7, after which you must have exactly 7 cars?
  19. Building cars

    I'm not understanding something about two shifts building exactly one car. A whole car (in one shift) or two half-cars (in two shifts) seem to be the only cases. Maybe spoiler one other possibility as a means of explaining? Thanks.
  20. Peter and Paul, who are neighbors, each threw a party last Friday. Bad scheduling, to be sure, but that's life. Even worse, their guest lists were identical: all 100 of their friends were sent invitations to both parties. When guests arrived, the happy sounds of those already present could be heard through the two open doors, and the old phrase "the more the merrier" figured in their choice of which party to attend: If at any point there were a people present at Peter's party and b people present at Paul's party, the next guest would join Peter with probability a/(a+b) and join Paul with probability b/(a+b). To illustrate: When the first guest arrived only the two hosts were present. (a = b =1.) So that choice was a tossup, and let's say that the first guest chose Peter's party. (a = 2; b =1.) Now the second guest would follow suit, with probability 2/3, or choose Paul's party, with probability 1/3. And so on, until all 100 guests arrived. What is the expected number of guests at the less-attended party?
  21. Jelly beans join the clean plate club

    It must be symmetric about the NW-SE diagonal, so your figure show all the cases you computed. Nice, btw. Hint
  22. Whodunit?

  23. Jelly beans join the clean plate club

    @plasmid Does the program imply a proof that it can always be done? Or is it a statement that no counterexample has yet been found? A proof could be a repeated procedure which after each application reduces the smallest number of beans on a plate. Does your algorithm always reduce the number on place C?