BrainDen.com - Brain Teasers  # harey

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## Everything posted by harey

1. 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
2. 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?
3. I hope I got it:
4. 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.
5. 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).
6. @hhh3: Can you tell how?
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.
12. 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.
13. 100% loss
14. Even here, it increases the travel time. A non-zero component of the plane's speed is used to compensate the wind.
15. This assumes that a point is (at least partially) illuminated if a straight line can be drawn uninterrupted between it and a light source.
16. 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.
17. 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.
18. 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?
19. Can you prove it? I hesitate between 1 and 2...
20. Well, I am just not very sure concerning the assumption in b). I do not consider whether there is a solution of another kind - when solving a problem, there is a little bit of intuition on the beginning. This does not make me doubt about the solution if I find one, except in the case I would claim it is unique. As for the remainder, there are so few possible solutions (we look for integers) that you can try them all. Otherwise, you could solve it as a system of multiple equations with multiple variables, but I think it is too much hassle here. I am awaiting the official solution!!!
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