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Everything posted by harey

  1. The speed is a real number. As there is an infinity or real numbers, p(two bullets have the same speed)=zero, at least for a "small" n. The same way we do not suppose that 3 (or more) bullets can collide. However, there is something else that gives me headaches. It works when there are n bullets in line AFTER all n bullets were shot. What if the first two bullets collide and then n-2 bullets are shot? More generally, what if there are collisions before all bullets are shot?
  2. Yes, the "official" one. But as I know it, it would be unfair to publish it. So I will leave it to someone else. We gave some hints now, did not we?
  3. The list of given examples is not exhaustive nor limiting. You can tailor a property to any number and we can argue ad eternum whether this property is interesting or not. Just my personal opinion
  4. Well, I know the "official" answer to this problem, but I do not agree with it. Human language is very imprecise, "interesting" is quite subject to interpretation. I.e. my ex does not find 4 interesting at all. 2+2=2*2=2^2? So what?
  5. I calculated very roughly. Interpolating an exponential is quite dangerous. If the temperature was rounded by 0.2 degrees and measured 2-3 minutes before/after the whole hour, you can get quite a different result. Proof is left to the reader.
  6. Not necessarily. v3>v2>v1 and diff_1<>diff_2 assumed diff_1=v2-v1 diff_2=v3-v2 if(diff_2>diff_1) then [3 reaches 2] else [2 reaches 1] I just cannot figure out p(diff_2>diff_1).
  7. What about a supplementary rule excluding Fred has 354?
  8. I have got 3 solutions, but I do not really understand the point (9)
  9. I calculated with 660.60 and the price is 600.60. Just a small typo as I often do...
  10. Set aside the sets calculated by Bonanova: 13,519.90-(4*660.60)-(1*1501.50)=9,376 As 600.60*5=1501.50*2=3003=3*1001, remain TVs for 231, 273, 429, 1001 (multiple solutions reserved). 231=3*7*11 273=3*7*13 429=3*11*13 -> all divisible by 3 9+3+7+6=25; so 9,376 is not divisible by 3; -> there are at most 8 TVs priced 1,001 and only 7,374 and 1,368 fits. The same way substract multiples of 231 and consider multiples of 39 (3*13, common divisor of 273 and 429). Alternatively, I think the problem could be solved by Diophantine equations, but I never studied them.
  11. I do not follow here. Worst case: the cards are sorted. Bonanova wrote: What is the expected number of cards in the left pile after all N cards have been drawn? OK, that answer is not difficult to determine, since the definitions of the piles are symmetrical.
  12. Even here, it increases the travel time. A non-zero component of the plane's speed is used to compensate the wind.
  13. This assumes that a point is (at least partially) illuminated if a straight line can be drawn uninterrupted between it and a light source.
  14. These thoughts suggest what I think is an interesting question: for each N, find an example of a convex solid that can be illuminated with the fewest lights. That's the start to the original question What bothers me with points and lines: We do not illuminate an object as such, we illuminate it's surfaces. (Already a problem with a sphere and alike: in this case, we may not consider the sphere has one surface, but infinity.) If a surface is illuminated, does it imply that it's edges and vertexes are illuminated, too? One answer would be that this question does not have a sense, but it seems a little week as argument.
  15. Not so quickly... If we need (in 3D) four lights for a sphere, a cylinder can be illuminated by three lights only. (I spent a long time to find a way to show that the edges can be smoothened and still illuminated until I consulted http://en.wikipedia.org/wiki/Convex_body and realized that a cylinder is a convex solid). For a cone, two lights are enough. What bothers me at most: Consider a unit cube in the xyz coordinates and a line (0;0;0) - (1;1;1). If we need 0 lights to illuminate a point or a line, the whole cube would be illuminated by just two lights placed on this line just slightly outside the cube.
  16. c - a = -1 You mistakenly have this as positive one in your post. This is just a typo, sorry for that: d) ac=2; (c-a)=1; I would find very tasteful a=2; c=1; (1-2)=-1 I'm looking for you to show the steps in solving the system of simultaneous equations above. Sorry, I fear I will disappoint you (as I already said, there are few possibilities, so a brute force approach will solve the system in seconds while I would spend hours due to algebra errors): for a in [-2,-1,1,2]: c=int(2/a) for b in [-2,-1,1,2]: d=int(-2/b) if(d-b==3): if(a*d+b*c==3): print("a=",a,"c=",c,"b=",b,"d=",d,"ad+bc=",a*d+b*c,"d-b=",d-b) print(a*c,"x2+",a*d+b*c,"xy+",b*d,"y2+",c-a,"x+",d-b,"y-1") The problem does not have to be solved by a system of equations, As no one else posts and I have no other idea how to solve it, can you post the solution?
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