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bonanova

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Everything posted by bonanova

  1. Al, Bert, and Charlie competed in a track and field event in which points were awarded for 1st, 2nd, and 3rd, place only. At the end of the day, Al had accumulated 22 points, while Bert and Charlie each garnered only 9 points. No other competitor earned points. Bert was 1st in the shot put. Who finished 2nd in the javelin throw? This is a Gold star puzzle.
  2. Sir, I bear a rhyme excelling In mystic force and magic spelling Celestial sprites elucidate All my own striving can't relate. See, I have a rhyme assisting My feeble brain, its task ofttimes resisting.
  3. You are given 5 circles, A, B, C, D, and E, whose radii are, respectively, 5", 4", 2", 2", and 1". Can you find a way to overlap circle A with portions of some or all of the other four circles so that the un-overlapped portion of A has the same area as the sum of the unoverlapped portions of the other four circles? That is, the red area is equal to the sum of the green areas. Circles B, C, D and E may overlap portions of each other as well as a portion of A.
  4. Find a, b, c. ab x ca= abca , a 4-digit number
  5. Here’s a challenge: I think this puzzle can be solved with almost no math at all. Like, think of a single “what if” question to ask that changes the conditions slightly.
  6. I have a strong feeling, and I'm working on a proof, that
  7. Which way should we evaluate this? [ [ [ [ [ x^(x/2) ]^(x/4) ]^(x/8) ]^(x/16) ]^(x/32) ].... or x^[ (x/2)^[ (x/4)^[ (x/8)^[ (x/16)^[ (x/32) .... ] ] ] ] ]
  8. One approach, among several, is to and
  9. I guess I've always been a little confused about what is being asked. Sharing is not permitted, yet an "average" result is requested. If average values are used, then two holes is enough. If there is no sharing or averaging, survival is assured only by the (very unlikely) worst case of 100 holes. If the question is what is the expected number of holes that together yields at least 100 nuts, we have an answer from simulation. Is there a way to say precisely what else might be needed?
  10. @harey, some guidance about completing a formula would be appreciated. I'm interested in the "surprise."
  11. Yes you cand do that. The elevation of the two men as they ascend and descend are continuous functions of time. If at one time one is greater and at a later time the other is greater there must be a time when they are equal. Since they both stay on the path, and if the the slope of the path never changes sign, then they meet at that time. If the slope does change sign, then multiple points on the path will have the same elevation and there can be multiple times when the elevation of the two men are the same. They meet at one of those times.
  12. Good start -- That's shorter than 3, which comes from any three of the square's edges.
  13. Al and Bert are among 100 passengers assigned to one hundred seats on an airplane. Al was first to board, and Bert was last. Strangely, the first 99 passengers ignored their boarding passes and took random unoccupied seats. Bert liked the seat he was assigned and is not happy with the situation. If he's lucky, his seat is unoccupied and there's no problem. Otherwise, he insists the passenger erroneously occupying it move to his own assigned seat. The displaced passenger must then move, possibly displacing another person. This process continues until all passengers are seated. What is the probability that Al must move?
  14. Recently we considered the shortest roadway that connects the four corners of a square. Here we seek the shortest set of line segments, one attached to each of a square's corners, that need not connect with each other. Instead, what we ask of the line segments is that is they will block any ray of light attempting to pass through the square.
  15. Consider a random walk in the plane where each step is taken, beginning at the origin, in either in the positive x or positive y direction, i.e. either east or north, each choice being made by the flip of a fair coin. The length of each step is 1/2 the length of the previous step, and the first step has length √2. After infinitely many steps have been taken, what is your expected distance from the origin? Edit: Ignore the original text in pink. Instead, What is the distance to the origin of the centroid of the possible termination points? You find the centroid of a set of points by averaging respectively their x- and y- coordinates. First correct answer wins, but style points will be awarded as well.
  16. @plasmid Nice solve. The puzzle itself was more a math exercise than a puzzle -- that part was thinking it through and setting up the calculation. I thought is was interesting in that you can sort of envision the setup and know that it had to be e.g. greater than 1/3, but maybe not as great as 2/3, so the question was would it be greater than 1/2?
  17. Yes there is a rule, L'Hopital's rule. Basically you can just replace functions by their derivatives to resolve indeterminate values. Or, you can just evaluate expressions and see how they behave: It's intuitive that the exponential dominates for large x if you think of say x / e3x, instead of x e-3x.
  18. I guess we can compute expectation value as well:
  19. The King has decreed that his daughter the Princess shall marry the most wonderful Prince in all the land. One hundred suitors have been selected from their written applications, and on a certain day the King arranges for them, in turn, to interview the Princess. Each suitor must either be chosen or eliminated on the spot. If the Princess does not choose any of them, she will marry the last Prince to speak with her. You have been chosen as the Royal Advisor to the Princess and tasked with implementing her best strategy to choose the Most Wonderful Prince of the realm. You devise an evaluation scheme by which the princess can assign a unique "wonder number" to each prince as she meets him. The strategy then is to have the Princess reject, but record the highest score of, the first N princes that she meets. The Princess will then choose the first Prince that she subsequently interviews whose score exceeds that recorded score. That's it. The puzzle is basically solved. Except, of course, to decide on the optimal value of N. It requires some thought. If N to too high, the most wonderful prince is likely to be eliminated at the outset, and she ends up with the last guy. If N is too small, the Princess will likely settle for a fairly undistinguished prince. What value of N optimally balances these two risks? What is the probability that the Most Wonderful Prince will be chosen? Disclaimer: I recall this puzzle being posted before, with different flavor text. And it's somewhat of a classic. To give it a fair play here, I'll ask not to post any links and not to just give the answer if you know it, at least not without "showing your work."
  20. There is survival in numbers.
  21. The cars in front of the slowest car are the remaining cars. (Each slowest car captures those behind it.)
  22. I think this puzzle is exactly the same as the traffic jam puzzle I posted recently, although that's not obvious at first glance. @plainglazed found a solution in which he formed clusters of cars by recursively locating the slowest of a group of cars, assuming on average it was in the center of the remaining cars. This corresponds to "grabbing" on average one-half of the remaining marbles from the bag. Picture the marbles in a line and, grabbing a random percentage of them starting from one end. This has to end up having the same number of marble grabs and car clusters. It leads to a logarithmic answer, but to the wrong base -- it should be the natural loge, not log2, which gives too large an answer. Instead of decreasing the number by 1/2 each grab, the remaining number is decreased by 1/e each grab. Same must go for locating the slowest of the remaining cars. @plasmid found a solution that leads to Sum { 1/k }, which as you point out is ln { n } + gamma, and is here confirmed by simulation. What I can't find is a corresponding analysis for the marbles problem (although plainglazed's approach is applicable to both problems) that will lead to that same sum, instead of yours, where both sums appear to be correct. This has been fun to think about. .
  23. That agrees with simulations I ran, which gave, approximately, Nice solve.
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