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  1. Today
  2. Worth the Weight

    4 light ones/48
  3. @CaptainEd brilliantly answered a puzzle that I mis-worded into a much tougher one. Nice job. Here's the puzzle I had intended to post: You have just lost your 143rd straight game of checkers and have vowed never to play another game. To confirm your vow you decide to saw your wooden checkerboard into pieces that contain no more than a single (red or black) square. With each use of the saw you may pick up a piece of the board and make one straight cut, along boundaries of individual squares, completely through to the other side. You wish to inflict as much damage as possible with each cut, so you first calculate the minimum number of saw cuts needed to finish the job. And that number is ... (spoilers appreciated.)
  4. When midnight strikes

    Isn't this fun? The { real numbers } between 0 and 1 comprise two infinite groups: { rationals } and { irrationals }. Rationals (expressible as p/q where p and q are integers) are countably infinite. We can order them, by placing them in an infinite square of p-q space and drawing a serpentine diagonal path. But the irrationals are not countable. The cardinality (notion of size used for infinite sets) of the rationals is called Aleph0 and for the irrationals (and reals) it's called Aleph1 or C (for continuum.) So first off, what we can do with the { numbers } between 0 and 1 depends on { which numbers } we want to deal with. Next, there's a problem that points are not objects that can be moved from place to place, as coins can. Points are more descriptions of space than of autonomous objects. If 0 and 1 are on a number line, the location midway between them is the point denoted by 0.5. It can't be removed. (We could erase the line, I guess, but that would not produce an interval [0 1] devoid of numbers, either.) But we can finesse this matter by "painting" a point. Kind of like what we did when we inscribed a number on each coin. Painting does a little less - it puts a point into a particular group or class but it does not distinguish among them. We can tell coin 2 from coin 3 and also distinguish both from a coin with no inscription. Two blue points, however, are distinguishable from unpainted points, but not from each other. But for our purpose here, painting is actually enough. Since the { rationals } are countable, we can paint them, in sequence, and if we do that at times of 1, 1/2, 1/4 .... minutes before midnight, all the rationals between 0 and 1 will be painted blue by midnight. (I don't know of a similar scheme for completing uncountable task sets, so the irrationals alas must remain unpainted.) Next, instead of asking whether all the points in [0 1] were removed, we can ask, with pretty much the same meaning, whether at this point the entire interval [0 1] has been painted blue. It should be, right? After all, between any two rationals lie countably infinite other rationals. So the paint must have covered the entire interval, right?. Well ... actually ... the answer is counter-intuitively No. And this is because even though the { rationals } are "dense" meaning there are no "gaps" between them, as noted above, the { irrationals } are even "more dense." That is, between any two irrational numbers there are uncountably many other irrationals. That means, for every blue point we created in the interval [0 1] there are uncountably many unpainted points. We can't magnify the line greatly enough to "see" individual points, but "if we could," if we were lucky enough to come across a single blue point we'd have to pan our camera over uncountably many unpainted points before we saw another blue one. In measure theory, the measure of the rationals over any interval is zero. The measure of the irrationals over [0 1] is unity. So what of our quest of emptying [0 1] of points? It's the same as our quest to paint [0 1] blue by painting all the rational numbers in that interval. (Reminding ourselves that there were too many irrationals to paint one at a time.) Turning again to measure theory, the measure of the "blueness" of the interval would be zero. That means we would not see even the hint of a faint blue haze. Back to our "empty the interval by removing the rationals" quest, Nope. [0 1] would not be empty: uncountably many points (numbers) would remain.
  5. Right Prime

    Some amount of primes.
  6. Yesterday
  7. Right Prime

    Say 5939 is a "right" prime because it remains prime after dropping any number of digits from the right: 5939, 593, 59, and 5 are all prime. How many right primes are there less than 1000?
  8. Four dogs are positioned at the corners of a square (d=1m), chase each other in clockwise direction with the same constant speed . As their target is moving, they will follow a curved path, eventually colliding in the center of the square. Why is the total length of the path just 1m?
  9. Worth the Weight

    I think I've got it. Nah, I definitely don't have it. I can't account for one particular case.
  10. 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?
  11. Worth the Weight

    This is a nice puzzle.
  12. 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.
  13. 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.
  14. 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.
  15. 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?
  16. 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?
  17. 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.
  18. 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.
  19. Last week
  20. 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
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