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  1. Today
  2. Worth the Weight

    I think I've got it. Nah, I definitely don't have it. I can't account for one particular case.
  3. When midnight strikes

    Maybe an idea. There is an infinite number of numbers between 0 and 1. I have the opportunity to discard an infinite number of numbers between 0 and 1. How much are you willing to bet that nothing remains between 0 and 1?
  4. Worth the Weight

    This is a nice puzzle.
  5. When midnight strikes

    That is what infinity does to our brains. Al retains coin 2 at step 1. But Al discards coin 2 at step 2. What is true for coin 2 is true for every coin. Every coin has a scheduled pre-midnight discard date. So "what happened to the N coins not discarded?" If N is finite, then it's not midnight yet, and the box does in fact contain coins. We have to be patient. The process has to run its course. And, specifically, {coin n+1} will be discarded at step n+1, at time t n+1, prior to midnight. So after midnight, it will be gone. Along with all the others. Every coin, identified by the number engraved on it, has a well-defined pre-midnight discard date. But then if we're Bert, who never schedules the discard of an odd coin, we'll have a ton of coins.
  6. When midnight strikes

    I would dispute your first point. I can't immediately think of a scenario that permits a discard which will not eventually happen given an infinite number of opportunities. Can you provide one? We agree on the second point. My previous post answers precisely that question.
  7. Destroying a checkerboard

    @CaptainEd Wow. Not the answer I had in mind -- a much better one! Nice. Bonus (slightly modified) version. Same question, but this time no cut may end short of the opposite edge. It must go entirely through the piece being cut.
  8. There is a hotel with an infinite number of rooms, all rooms occupied by little green men (one man in a room). An infinity of little blue man arrive and each one asks for a room. No problem, the manager moves the blue man from the room n to the room 2*n, freeing the odd-numbered rooms for the green men. So far, loosely copied from https://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel. It turns out that the blue men sing between noon and midnight and sleep between midnight and noon while the green men sleep between noon and midnight and sing between midnight and noon. Complaints. The manager decides to group them. Conveniently, the rooms are in a straight line, numbered from left to right. While there is a green man left to a blue man, he makes them change the rooms. Eventually, all the blue men leave. - how many rooms are free? - how many rooms remain occupied? - what is the number of the first occupied room?
  9. When midnight strikes

    If a coin can be discarded, it does not mean it will be discarded. I would reformulate it: The key question is this: will all coins that are kept at a certain event ever be discarded at a later event? BTW, we can establish a bijection between Al's and Bert's coins. The coins bear green numbers. After each step, Al renumbers them and assigns them blue numbers 1, 3, 5, ... His blue numbers will match the (green) numbers in Bert's box. For any number of steps. I think it is legal to assume it is true even for N-> inf. Another way to prove that Bert's box will not be empty: graphical presentation. The number of coins in his box is a straight line (at 45 degrees). How is that it suddenly drops to 0? And maybe a corollary: Bert never discards more coins that he receives. How is that when he has, let's say 8 coins, he can have less in a later stage? If we reason with {coins} and {events}, don't you see a 1-1 relation?
  10. Worth the Weight

    hey bn - Thanks for getting this one started. You are correct in all you say above including your initial summary of the problem. Have edited the OP to include that explanation. I assure you there is a scheme in which one can guarantee finding seven heavy coins. Perhaps start off by finding one guaranteed heavy coin in two uses of a balance scale. Or not. I have no doubt you or others here will discover the method for finding seven heavy dimes.
  11. Worth the Weight

  12. When midnight strikes

    @harey Hilbert's hotel tells us not to treat countably infinite sets the same way we treat finite sets. There is no 1-1 correspondence between { 1 3 5 7 9 } and { 1 2 3 4 5 6 7 8 9 10 } as there is between the (infinite set of) { odd integers } and the (infinite set of) { integers.} Hilbert's hotel always has room for more. Completion (of an infinite set of tasks) is another tricky concept. If we number some tasks 1 2 3 ... and there is no final integer, how can there be a final task? And if there's no final task how can we complete infinitely many tasks, or describe the state of things after they have transpired? We finesse that point with a 1-1 correspondence of event times to the terms of an infinite series, 1 1/2 1/4 1/8 ... 1/2^n ...., which (conveniently) converges. We pack a countably infinite set of events into a time interval that ends at midnight. "Completion" is cleverly accomplished in two seconds. It's non-physical enough that it never could happen, but we can reasonably discuss the post-midnight state of affairs nonetheless. We have two countably infinite sets, {coins} and {events}. At each event, two coins are added to a box and one coin from that box is discarded. So, of the two added coins, at least one is kept. We know that at midnight the entire set {coins} has entered each box, and some (proper or not) subset of them has been discarded. The key question is this: can a coin that is kept at a certain event ever be discarded at a later event? For Al, the answer is yes. Al always discards his lowest coin, so at each event time ti he discards coin ci. Thus eventually that is, upon completion of the infinite set of tasks, every coin that is initially kept is later discarded. At midnight no coins remain. Al's box is empty. For Bert the answer is no. Bert discards the highest-numbered of his coins, and that is always the even coin that he just received. At no event is Bert ever scheduled to discard an odd coin. Every odd coin that enters Bert's box is kept, and it stays there forever. Bert's box contains a countably infinite set of coins. For Charlie the answer is ... well ... um ... actually ... I guess ... yeah, but it might take an infinite number of events for it to be discarded. Well it just so happens that we have an infinite number of events that follow the keeping of every one of Charlie's initially kept coins. So, yes. All of Charlie's coins that are not immediately discarded are eventually discarded. At midnight, Charlie's box is empty.
  13. Yesterday
  14. Worth the Weight

    You are shown a pile of dimes all of which have one of two distinct weights differing by a small amount not detectable by feel. Forty eight dimes are separated from this pile and you are told of these forty eight, light ones are a dime a dozen (literally - i.e. 44 heavy dimes and 4 light dimes). Using a balance scale twice, find seven heavy dimes. EDIT: for clarification
  15. Destroying a checkerboard

    That would not be permitted. The OP says with each use of the saw you may pick up one piece of the board and make a straight cut.
  16. Last week
  17. Baggage on a conveyor belt

    Here’s my attempt at an analytical solution
  18. Increase the square

    Nevermind...misunderstood the OP (again).
  19. What Question Must Be Asked?

    I was so stuck in my solution that is exactly the same as yours except that the exclusion came earlier that I reacted too quickly. Next time, I will read more carefully, promised
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