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  1. Past hour
  2. When midnight strikes

  3. Today
  4. 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.
  5. Yesterday
  6. Baggage on a conveyor belt

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

    Nevermind...misunderstood the OP (again).
  8. 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
  9. When midnight strikes

    After N steps, they will have received 2*N coins and withdrawn N coins. At that moment, there will be 2*N-N coins in the box. If the box is empty at midnight, this implies: limit(2*N-N)(for N->inf) = 0 At least a little bit surprising. @ThunderCloud I have some troubles to refute your argument. If you remove an infinity of finite numbers from infinity of finite numbers, it does not imply no finite number remain. (Not sure I am convincing and clear enough.) Counterargument: Al removed all coins 1 - N, coins > N remain. If N -> inf, numbering looses it's sense, but he did not remove all coins. As for Charlie, I am ruminating, too. The first idea: every number will remain with p=1/2. Wrong, 1 will be more likely removed than 99. 2nd idea: 1st step, 2 coins: p(removing 1)=1/2 2nd step, 3 coins: p(removing 1)=p(1 was not removed in the first step) * 1/3 = 1/2 * 1/3 = 1/6, p(1 remaining after 2nd step)=1 - 1/2 - 1/6 = 1/3 3rd step, 4 coins: p(removing 1)=p(1 not yet removed) * 1/4 = 1/3 * 1/4 = 1/12, p(1 remaining after 3rd step)=1 - 1/2 - 1/6 - 1/12 = I will not venture further, but this will not converge to 0. (Compare to 1 - 1/2 - 1/4 - 1/8...)
  10. Destroying a checkerboard

    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. 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.)
  11. Baggage on a conveyor belt

    Gah! Mine was not a useful clue. The denominator (greater than 6 but not huge) is too large for a simulation to tell you the fraction. Also I think your simulation value is high. But you're thinking in the right direction. Here may be more useful clues.
  12. Increase the square

    Four pegs begin at the corners of a unit square on a grid having integer coordinates. At any time one peg may jump a second peg along any straight line and land an equal distance on its other side. The jumped peg remains in place. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + O + + + + + + O O + ==> + + + O O + + + + O O + + + + O + + + + + + + + + + + + + + Is it possible to maneuver the pegs to the corners of a larger square? + + + + + + + O + O + + + + + + + + + O + O + + + + + + + + + + + + + +
  13. Last week
  14. Baggage on a conveyor belt

    Ok I tried using random numbers, got a fraction...
  15. Pairing the points

    Nice. A construction is certainly a proof of existence. A pairing without intersects exists and ... here it is! Now I wonder if for any groups of n blue and n red points there is only one pairing without crossings?
  16. Pairing the points

    Probably not what you have in mind, but I realized part of my previous answer was superfluous so this is a little more elegant and gets rid of some hand waving.
  17. What Question Must Be Asked?

    Proof of my claim...
  18. What Question Must Be Asked?

    Better still...
  19. What Question Must Be Asked?

    Assuming we need to find the number in each cell, here’s a variant that will give the others a bit more information: “is the total either 9 or 21?” this shows the others that Kleene has 4, 5, or 6. I don’t see that that’s enough yet. But maybe y’all do.
  20. When midnight strikes

    Thanks Thundercloud, that makes sense. Ok I’m on board with two empty boxes.
  21. When midnight strikes

    Is Confucious still alive and well?
  22. When midnight strikes

    Ok I believe Al is empty...
  23. When midnight strikes

    But...but...but, the OP points out that after N events, there will be N coins in the boxes! I always found infinity confusing, but how will they become empty?
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