harey

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About harey

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  • Birthday February 26

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harey's Activity

  1. harey added an answer to a question Badminton   

  2. harey added an answer to a question Half-balls   

    @hhh3: Can you tell how?
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  3. harey added an answer to a question numbers by the book   

    What about a supplementary rule excluding Fred has 354?
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  4. harey added an answer to a question numbers by the book   

    I have got 3 solutions, but I do not really understand the point (9)

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  5. harey added an answer to a question Can't find the receipt   

    I calculated with 660.60 and the price is 600.60.
    Just a small typo as I often do...
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  6. harey added an answer to a question Can't find the receipt   

    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.
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  7. harey added an answer to a question Can't find the receipt   

    What about:

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  8. harey added an answer to a question Highest of the lows   

    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.
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  9. harey added an answer to a question Relatively prime   

    100% loss



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  10. harey added an answer to a question Airplane with or without wind   

    Even here, it increases the travel time.
     
    A non-zero component of the plane's speed is used to compensate the wind.
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  11. harey added an answer to a question Illuminating a Convex Solid At Higher Dimensions   

    This assumes that a point is (at least partially) illuminated if a straight line can be drawn uninterrupted between it and a light source.
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  12. harey added an answer to a question Sides = Diagonal   

  13. harey added an answer to a question Illuminating a Convex Solid At Higher Dimensions   



    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.
     
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  14. harey added an answer to a question Illuminating a Convex Solid At Higher Dimensions   

    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.
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  15. harey added an answer to a question Maybe factoring a polynomial completely   

    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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