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harey

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Everything posted by harey

  1. Bonus question for those who are at ease with formulae: Using this proceeding, in how many parts can you cut the cake at most?
  2. What does mean "randomly"? If I remember well, there is a controversy how to choose randomly points on a circle.
  3. There is: just the formula giving the number of boys. But it is quite unreadable and to understand it, you have more or less to go thru what I have written.
  4. There can be a solution in one pass disregarding the answers of C. The numbers are a little bit high for practical purposes, so I will illustrate it on a sum of 12 or 13, everyone having 4. Notation: "!n" means "I do not see the number n" "<-(xy)" means "because otherwise I would know the combination is (a=x b=y)" Both A and B see c=4, they both know a+b=8 or a+b=9. They also know that the other one knows that. By saying "I do not know", they in fact tell: 1 A: !8 <- (18) B: !8 <- (81) !1 <- (17); (18) excluded in 1A 2 A: !1 <- (71); (81) excluded in 1B !7 <- (27); (17) excluded in 1B B: !7 <- (72); (71) excluded in 2A !2 <- (26); (27) excluded in 2A 3 A: !2 <- (62); (72) excluded in 2B !6 <- (36); (26) excluded in 2B B: !6 <- (63); (62) excluded in 3A !3 <- (35); (36) excluded in 3A 4 A: !3 <- (53); (63) excluded in 3B !5 <- (45); (35) excluded in 3B --> B knows A sees 4. --> A realises he never will get more information A already knows that b=4, he can imagine this dialogue (the answers of B will not change whether a=4 or a=5), so his very first answer is: 1 A: I will never be able to tell. B thinks: if b=4, this makes sense if b=5, A would have announced he would be able to tell. From the answers of A and B, C deduces c=4 (if necessary, I will explain). The remaining problem is whether the answers of C are redundant or whether they lead to another distribution. Comments?
  5. There might be a trap. Back to the secondary school for the definition of the measurement error. I have rulers which are divided in cm and subdivided in 1/10 of cm. If I tell you I mesured 2.4 cm, you assume the the value is between 2.35 and 2.45. Error in percentace=0.1/2.4*100 If I tell you I mesured 7.4 cm, you assume the the value is between 7.35 and 7.45. Error in percentace=0.1/7.4*100 I happen to have a curiosity of a ruler in inches. It's subdivisions are 6 (for 6 lines per inch or 12 cpi), 8 (for 8 lines per inch or 8 cpi) and 10 (for 10 cpi). (Supposing this ruler shrinked by 1/24, I get a headache.) If antelOpe has such a ruler, the error is 1/subdivison/4*100.
  6. A priori, the p of any card being red is the same, so it does not matter on which card you bet. (Therefore the best strategy is to bet on the first card because you are not losing time.) Roulette gamblers will not agree with this, they will argue that if there remain 3 red cards and 1 black card, p=3/4. Is there an easy way to calculate the p that R red cards and B black cards remain?
  7. Well, I have never been to LA, but I know what happens in Germany. Bus driving 80 km/h, behind a car driving 90 km/h that begins to overtake without caring much about the cars arriving at 160-260 km/h. When it comes to a stop, I often observed that the first pack of 5 cars needs 13 seconds to move, the second pack 12, the third 11 and the remaining 10. As if the drivers on the begin would need more time to take the decision.
  8. There is no mathematical reason for that, yet you are right. Any idea why?
  9. I started with the left side. IMHO, the only symbols that can be reversed are operators (I do not believe i.e. that -8 is reversed 8). Just that there cannot be an operator alone. Any other idea?
  10. Wording made some of the rules sort of shaky.
  11. Now, I have a problem. witzar was the first to answer and his answer is correct. ThunderCloud answered 1 minute later and his explication is clearer. Can I attribute two Best Answers? Anyway, congratulations to both that they did not fell into the trap.
  12. Let's toss a fair coin and note the events. When H H H appears, I pay three bucks When T H H appears, you pay one buck. And then we start over. Who is willing to play?
  13. Yes, I see. There must be a lot of solutions, like taking a sphere and a cube and drilling holes into the cube until it becames lighter than the sphere - no formula, no misscalculation. My preferred solution is still that of the three squares and three pankaces in the planes xy, xz and yz. BTW, the Volume=3a2d-d3 - I have the same oversight in my post n. 8. I just wonder how often we make false deductions in the real life in the style "three projections=circle => it must be a sphere".
  14. The best way is to make them of equal volume ( suppose what I call a "3D square" is a prism?): v_prism=a*a*d_prism v_cylinder=pi*a/2*a/2*d_cylinder => d_cylinder=(4/pi)*d_prism While it might work for small d, with growing d, it will became more and more perceptible that the cylinder should be a kind of barrel (or a pancake). The formulae get too complicated for my taste and my possibilities.
  15. The circle is inscribed to the square... That's easy. The circle is inscribed to the square.
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