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bonanova

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Everything posted by bonanova

  1. bonanova

    UN Mafia III

    Sad to see Araver gone as his analyses are usually pretty insightful, and I rank myself still as a newbie. Anyway that said, I'm fairly certain that y'all want me in the game for reasons stated in a previous post. Hello MikeD: Players: 1. Thalia - voting for bonanova 2. bonanova - voting for MikeD 3. Molly Mae - voting for Thalia 4. Flamebirde - voting for bonanova 5. maurice - quarantined by Sweden 6. MikeD 7. plasmid - Invaded N1 by China
  2. I would cast a mild vote for G / B / Indie game because I'm such a novice. Having said that, the present rules are nicely done.
  3. I'll jump in. (Host): araver Players: 1.Thalia 2.bonanova 3. 4. 5. 6. 7. 8. 9. 10. 11. Backups 1. 2.
  4. Hi Kt.Kpop. Welcome to the Den, and congratulations on solving the puzzle. Feel free to post some of your favorites here, too. Interesting twist phaze. Is this an accurate paraphrase? Arrange 2,4,5,7,8,11 (differently) on three dice in such a manner that throwing each of them once cannot disclose enough information to deduce the six numbers separately. My intuition says that it is possible. But proving it seems to require a list of every combination of all possible throws, 24 possibilities for each throw, and the application of a program that is verified to find a solution if one exists for each set of three throws. Proving something does not exist is always harder. Still it's interesting.
  5. Well the issue is what 0,000...1 means. That question must be answered before asking whether it has a value, and if so what the value is. The expression contains a string that is both infinite and terminating. A terminating string can be counted. Each digit occupies a numbered place. What place does the 1 occupy? This expression implies that a finite string that terminates with a 1 can contain an infinite sub-string of zeros. That fact is equivalent to saying that any explanation of the meaning of the expression must contain a contradiction. We can construct the sequence 0.1 0.01 0.001 0.0001 0.00001 ... and ask whether the sequence converges. Yes. It easily can be proved to converge to zero. But the proof never mentions 0.000...1. So it does not apply to the OP. Regarding your student's proof I question the assertion 1 - 0.000....1 = 0.999.... Why? Because it equates an infinite string to a terminating string. I would say 1 - 0.000....1 = 0.999... - 0.000...1 = 0.999...8 whose meaning is equally problematical. Saying that it evaluates to unity carries the same difficulty as the original question.
  6. An airplane was about to crash. There were four passengers on board but only three parachutes. The first passenger said, "I am Steph Curry, the best NBA basketball player. The Warriors and my millions of fans need me and I can't afford to die." So he took the first pack and jumped out of the plane. The second passenger, Donald Trump, said, "I am the newly-elected US president and I am the smartest president in American history, so my people don't want me to die." He took the second pack and jumped out of the plane. The third passenger, the Pope, said to the fourth passenger, a 10-year old school boy, "My son, I am old and don't have many years left. You have more years ahead so I will sacrifice my life and let you have the last parachute." The little boy said, "That's okay, Your Holiness, there's a parachute left for you. America's smartest president took my school bag."
  7. Well the author has now been outed. Suspect others had it as well. But the message still has to be cracked. Think logically. Good job.
  8. Must have been nice to have a unique solution appear like that. I'm working on a truth-teller puzzle based on the recent US election. Going through the same process.
  9. It was surprising to find that there is only one solution. Here's how I found it. It's computationally challenging - brute force is possible, but I took it as a challenge to solve it efficiently. Even so, aside from guessing, there is no way to avoid evaluating the expression ((AoB)@C)%D for all 9x8x7x6=3024 variable permutations and all 6x5x4=120 operator permutations, 362880 computations in all. But I only did each once. Next I noticed that not all the results were valid. It turns out there are only 3928 valid expressions - about 1.1% of the cases. But all 362880 have to be evaluated to determine that fact and find the valid ones. Each valid expression represents a candidate for a side of the triangle. I put the results into a table. No further calculations were necessary. The first 15 rows (candidate sides) look like this: A solution to the triangle is a set of three sides that "Connect" at their end values (and ops) Use all the values Use all the ops Share a common expression value (equal to the product of the three corners) There were 60 valid results for the expressions: And the full table did not have to be searched. Only segments of the full table had to be searched: those records with a common result. For example, the 2 sides that evaluated to 240. Or the 13 sides that evaluate to 252. Or the 79 sides that evaluate to 45. That is the only solution.
  10. Not surprised. I wondered whether this belonged in the riddles section. It's both that and logic. Happy belated thanksgiving.
  11. Around World Series time recently, (that's the American baseball championships for our international Denizens) I asked the question, dressed as a probability puzzle, "Given that you could pick one of the best-of-seven games in the series that you could guaranty to win, which game would that be? I even flavored the question with the opinions of experts that the pivotal games are often taken to be game 3 or game 5, as statistically these games correlated well with the eventual winner. And many responded with one of those choices for their answers. But the correct answer of course is game 7. (Or as some said, the Last game.) So I also like the "one with the car" choice.
  12. Adam, Bill and Charlie stood in a line with Adam in front and Charlie at the rear. Charlie could thus see both Adam and Bill, Bill could see Adam, but Adam could not see either Bill or Charlie.Their teacher brought out a box containing three red hats and two blue hats. She placed on each boy's head a hat from the box and asked them to name the color of the hat they were wearing. Although all three responded to the question, only Adam knew the color of his hat. And now so do you.
  13. Assuming "first" means lowest-numbered,
  14. Looks like Now I can get my life back. Nice puzzle.
  15. Nope. He is a "real" Bill, though. Just a little older than Clinton. Keep in mind that Picket's post is on the right track.
  16. You can use the Search feature here to see whether a puzzle might have been already posted. Mouse over Activity tab and click on Search. A version of Monty Hall has in fact been posted here, although with enough different terminology it's hard to find. Either way, a member might know of a previous post and simply direct discussion to it. Regarding usefulness, puzzles are usually posted without the solution. This permits members to post their responses to your puzzle. When you feel it has been solved you can click the check mark (upper left corner of their post) to so indicate. If it's not solved after some time has passed you might want to post a clue (putting the clue into a spoiler.) You can put anything you like into a spoiler so it's not visible unless a member chooses to see it. Simply highlight the part of your post to be hidden and click on the eye icon and then submit your post.
  17. Hi. That's a standard puzzle and you gave what I think is the best short answer
  18. Inspired by FUZZY's recent puzzle Suppose abcd is the normal decimal representation of a number: abcd = (1000a)+(100b)+(10c)+c. The digital sum of a number is defined as a+b+c+d. Are there numbers for which the product of its digital sum and its reversal is the original number? For example, the digital sum of 12345 is 15. Its reversal is 51. 15x51 is 765, 12345 is not equal ti 765, So 12345 is not a solution. Hint: 81 is a solution, as is 1. Are there others?
  19. Clarification? Each variable {A B C D E F G H I} can be used only once. I now assume, but it's not clear, that each operator {+ - x / ^ 10A+B]} can be used only once, making (9!)(6!) cases. I started allowing any operator to be used in each white triangle.
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