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.
I do not follow here. Worst case: the cards are sorted.
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.
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.
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.
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;
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]:
for b in [-2,-1,1,2]:
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?