BrainDen.com - Brain Teasers bushindo

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Everything posted by bushindo

1. Hopefully, this solves all tiebreak situations and removes the uncertainty. Reading from the OP: Given that it [Cole's strategy] is uncertain, what can we determine regarding Bobby's shooting accuracy? From your reasoning regarding Cole's strategy, what is the answer? I agree about semantics. I think the underlying crux of the discussion is the interpretation of 'uncertain'
2. Yes, that is the answer. Nice job. I believe the above solution conflates "trying to hit Alex" with "actually hitting Alex". Here is my reasoning
3. Indeed it should be a quadratic expression instead of cubic. My mistake. Good catch, Prime.
5. Assuming each participant is forced to shoot only at remaining duelists, then here's what I get Okay, so earlier we found that not allowing an airshot on Cole's turn reduces to a trivial solution. So, let's assume that on the first shot, Cole has the option of shooting A, B, or none of those two.
6. Your proposal (tethering 4 ships to the 5th) is akin to increasing the mass of the 5th ship by five times while also increasing the fuel tank five times. Your solution assumes that fuel usage for the navigating ship is the same whether it is alone or whether it has 5 times the mass. In physics, as in life, there is no such thing as a free lunch. You can't carry 5 times the mass to Mars without expending extra energy. Perhaps you are thinking that the navigating ship only has to set a course for the entire fleet, so the energy usage is independent of the mass. Recall that in space, if you need to change the course, you will need to change the momentum vector of the entire fleet. That means if you want to move 1 degree to the left, for example, the navigating ship will need to provide enough thrust (energy) to move its entire mass to the left that much. Even if we allow the conservation-of-energy violation, the 5 ships still can not make it to Mars and back. From your OP, "each spaceship has a fuel capacity to allow it to fly exactly 1/4 way to Mars" and "Each spaceship must have enough fuel to return safe to the base space station". Since each tank will take you 1/4 of the way to Mars, 5 tanks are enough for 1.25 of the distance to Mars. How are your 5 ships supposed to get to Mars and get home on 5 tanks?
7. Assuming each participant is forced to shoot only at remaining duelists, then here's what I get
8. Probably not what you were looking for, but here's what I recommend Cole do So, clarification please. Can any participant choose to shoot into the air instead of the other duelists?
9. Thanks for the kind words, bonanova, I also learned much from you over the years. The two wonderful examples (among many) that come quickest to mind are Hole in a Sphere and Maiden Escape, both of which taught me completely new ways of thinking.
10. You are both correct. bona_gold_star.gif nakulendu posted first, so I marked his answer as the solution. The solutions above are making some assumptions about the properties of underlying distribution of N, which may or may not be warranted depending on your interpretation of the OP.
11. It seems like this problem is dependent upon what is a reasonable a priori distribution for N. Oh wait, the answer is much simpler than that. Turns out this is the easiest bonanova puzzle I have ever seen =)
12. It seems like this problem is dependent upon what is a reasonable a priori distribution for N.
13. But where is the proof that it is the smallest number that meets the conditions? I can't prove that it is the smallest number. The solution that you posed made me realize that my solution is flawed. Originally, I thought I had a divisibility rule for 100 ones, which allowed me to compute the smallest integer within those rules. The number you came up with made me realize that the divisibility rule I had only worked on a subset of the possible divisors. So, there might be smaller solutions out there. I made a mistake in an earlier post. I should have called that number "the smallest number that I could find" instead of calling it the solution.
14. Excellent work so far. Sorry for not responding earlier. I was on a vacation and didn't have access to a computer Solution to the puzzle
15. Dear wolfgang, the solution that you consider correct suffers from two problems: 1) It violates common understanding of thermodynamics and 2) even if we allow such violation, it does not satisfy the conditions in your original post. Here is why While I understand that lateral and out-of-box thinking are valued around here, in general such solutions have to be internally consistent (and avoid violating explicit conditions set forth in the original post). Also, in this forum, we obey the law of thermodynamics.
16. So I guess the lateral solution of this puzzle is contingent upon the properties of this 'fuel'. My previous solution assumes that "the fuel spent is proportional to distance travelled". Since that doesn't seem to be the case, I'd like some clarification about the conditions of the puzzle. 1) My impression is that the fuel is used to make constant minor course corrections during the trip (per OP: "without fuel the spaceship will miss its direction") and to maintain the life-support and ship computer (per OP: "… and may explode"). Is this correct? If that is not correct, what is the fuel being used for? 2) Consider the following two scenarios * A single ship flies from base (x=0) to the one-fourth point (x=1/4) * A ship flies from the halfway point (x=1/2) to the three-fourth point (x=3/4) Does the ship use the same amount of fuel (1 tank) in both scenarios? 3) Are the ships supposed to land on the surface of Mars?
17. I don't think the number of ships can be reduced below 9. Here is why
18. If you want the number of unique ships required, then
19. I'm getting the max of Ha, silly me
20. k-man makes a good point. For the sake of the puzzle, let's assume that the fuel spent is proportional to distance travelled. Let's also assume that the ship that gets to Mars will need to return home.
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