Jump to content
BrainDen.com - Brain Teasers


  • Content Count

  • Joined

  • Last visited

  • Days Won


Everything posted by bonanova

  1. Yes, sir, that is exactly the right track.
  2. They are equally likely but they are not the same. But it's worse than that. The OP is deficient, because it does not tell us how we came to know what we know. Instead, let's create a situation where we know how we know what we know, and therefore will let us find the probability that "the other kid is a girl," unambiguously. Doing that, we know the answer is 1/3.
  3. So, while not being a correct solution, this would meet that qualification?
  4. Once letters start getting removed it quickly gets easier. At the start it's very hard. How much of a clue are you willing to share? I don't want to disclose too much, but would you be willing to confirm the politician is male and contemporaneous? Initials or country of affiliation might be too revealing.
  5. This sounds a lot, but not exactly, like eliminating variables from sets of equations. Is that the idea?
  6. Both answers state correctly that (a) winning distances give speed ratios and (b) combined speed ratio gives combined winning distance. What part of that can be more (or less) straightforward?
  7. At the annual Brain Denizen picnic, there were the inevitable games, among them the ever-popular three-legged race. Three teams were formed by tying one contestant's right leg to another' s left leg. Fortunately all six contestants made it to the finish line without any broken bones! For purposes of this puzzle we assume all three teams ran the 100-meter course at constant speed. Team 2, comprising BMAD and Thalia, were able to beat Team 3, comprising rocdocmac and DejMar by 20 meters, but lost to the winner, Team 1, comprising plasmid and plainglazed, by 20 meters as well. By how many meters did Team 1 beat Team 3?
  8. Very nice! There is another solution that ends two moves earlier, leaving pegs only on the edges.
  9. The OP does not say how many apples there are. It says the proportion that are poisonous. Question: was that the intent?
  10. You've probably seen this puzzle. There are 15 holes in triangular array. (See sketch below.) The game begins with pegs in 14 of the holes. The play is to jump pegs over adjacent pegs, removing the "jumped" pegs afterward, as in checkers. The jump is made in a straight line. To make a jump, you need a contiguous group consisting of { peg1, peg2, hole } in a straight line. Peg1 ends up in the hole, and peg2 is removed. The object is to make 13 legal jumps and end up with a single peg. This happens about 6% of the time. That is, about 94% of the time you get a configuration, with more than one peg remaining, that permits no further legal jumps. In some games the peg must end up in the original empty hole, and that happens only about 3% of the time. So, it's not a trivial puzzle. This puzzle asks for something different, and easier: Lose as badly as possible. That is, select a location for the empty hole, and then find a sequence of moves that leaves the greatest number of pegs on the board where there are no more legal jumps. It's simple enough to play, even without the game, by marking hole locations on a sheet of paper and using pennies. As already stated, there are 15 holes. There are also 36 possible jumps. For convenience in writing sequences of jumps, they can be numbered, as follows: Number the jumps like this: and the holes ---------------------------> o ------------- 1 So Jump #1 means the / \ like this: peg in hole #1 jumps 1 2 ----------> 2 3 over the peg in hole #2 into the empty 4 5 6 hole #4. o o / \ / \ 7 8 9 10 Jump #18 is peg 7 3 4 5 6 over peg 8, into 7 13 11 12 13 14 15 hole 9. / \ o-8 o 14-o Holes 4, 6, 13 / \ / \ / \ begin 4 jumps; 9 10 11 12 15 16 the others 17 19 21 23 begin two. / / \ \ o-18 o-20 22-o 24-o There are 36 jumps. 25 27 29 30 33 35 / / \ / \ \ o-26 o-28 31-o-32 34-o 36-o With symmetries taken into account, the holes have four equivalence classes: Corners (1, 11, 15) Adjacent to corners (2, 3, 7, 10, 12, 14) Edge centers (4, 6, 13) Centers (5, 8, 9) This means that there are just four distinct places for the empty hole to start a game: { 1 2 4 5 }. All other holes are symmetrically equivalent to one of these. Just to be sure the numbering above is understood, here is a winning game of the normal type. Start with pegs in every hole except #1. (The top hole is empty.) Then make these jumps: { 7 14 2 17 23 27 34 26 30 6 35 14 7 }. If done correctly, the original hole #1 contains the final peg. Enjoy.
  11. Doctoring the figure a bit (while I think about solving it.) o - o - o - o - o | / o o o o | / o - o - o Question: You mention the number of nails (12) and also the number of nails in the perimeter (here that's 10.) So for this example, would we be given N=12 or N=10? I'm thinking the latter, but would like to confirm.
  12. Welcome back TSLF. Nice puzzle!
  13. With that interpretation,
  14. Assuming the above, we note that
  15. Just to be clear, n = 3 4 5 7 8 9 11 12 13 ..., and CW and CCW alternate untethered (my new favorite word) to parity?
  • Create New...