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harey

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harey last won the day on January 2 2022

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  1. Here's an example from my code's 6 lantern output: 111110 <one lit lantern (interpret 0 as "on" and 1 as "off") 111101 <one lit lantern You suppose the devil turns the lantern 5 on. But what if he turns the lantern 4 and not 5 on? 111110 <one lit lantern (interpret 0 as "on" and 1 as "off") 111010 your turn P.S. I am not touchy (did not even think it could be interpreted this way), but I think things over. In the last time, I wrote a lot of nonsense, i.e. my first answer.
  2. @EventHorizonMy post is unrelated to yours. The number of lamps is not important, just a little bit more that those I quote. "Let's start with one lantern lit, i.e. 33.": This is not a binary number, it is the 33rd lantern assuming the starting position is the 1st lantern (conveniently chosen), following the lantern 32 and followed by the lantern 34.
  3. To decide who will do the dishes, we put in a jar tickets with prime numbers up to 111. The one who draws the ticket with a larger number wins. I look at my ticket: 3. Maybe you have a low number like 11... So I offer to exchange our tickets before showing them. What is the probability I will do the dishes?
  4. I get 13. Take one coin of each chest. a) You get two Gold, one Silver: Silver is identified, remains to distinguish all_Gold and Gold_Silver. Take 10 more coins from the first G: - if they are all gold, you identified all_Gold, the remaining is Gold_Silver - if you get a silver coin, you identified Gold_Silver, the remaining is all_Gold b) You get one Gold, two Silver: Same proceeding, permute Gold/Silver.
  5. A good question would be "How many steps on average until I win?"
  6. There is a strip of N (N>13) squares. In the middle is the SILVER DOLLAR. Two players alternatively place a penny on an empty square. Then, at each turn, a player must: - move one coin one or more squares to the left, observing: - the coin cannot move out of the strip - the coin cannot jump on another coin - the coin cannot jump over another coin - or - - pocket the leftmost coin Best strategy?
  7. @Plasmid I translated your notation into mine, as far as we are gone, we have same results (excepted some doublets). I thought about interpolation, too, but I did not go this way estimating there is not enough data.
  8. Same as rocdocmac Combinatorics:
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