harey
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harey added an answer to a question Badminton

harey added an answer to a question Halfballs
@hhh3: Can you tell how?

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

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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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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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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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harey added an answer to a question Can't find the receipt
What about:

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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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harey added an answer to a question Relatively prime
100% loss

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harey added an answer to a question Airplane with or without wind
Even here, it increases the travel time.
A nonzero component of the plane's speed is used to compensate the wind.

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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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harey added an answer to a question Sides = Diagonal

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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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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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; (ca)=1; I would find very tasteful a=2; c=1; (12)=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(db==3):
if(a*d+b*c==3):
print("a=",a,"c=",c,"b=",b,"d=",d,"ad+bc=",a*d+b*c,"db=",db)
print(a*c,"x2+",a*d+b*c,"xy+",b*d,"y2+",ca,"x+",db,"y1")
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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