BrainDen.com - Brain Teasers # bonanova

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## Everything posted by bonanova

1. One might start by constraining the points to be on the circle's perimeter.
2. ## jelly beans revisited - achieving equality

Did OP leave out the part about starting from a random distribution of the (total number of) b = 3x2n beans on the plates? Apologies.
3. ## Thank goodness, no remainder is chasing me

Close, check the second case. Brain fart. Nice solution.
4. You're correct. It can be a little bit shorter than that.
5. ## Thank goodness, no remainder is chasing me

Ten years ago I called attention to a number that when divided by a single integer p it left a remainder of p-1. (Help, a remainder is chasing me) Here is a chance to construct a nine-digit number, a permutation of { 1 2 3 4 5 6 7 8 9 } that has no remainders, sort of. The task is to permute { 1 2 3 4 5 6 7 8 9 } to create a number whose first n digits is a multiple of n for any single-digit n. For example, consider 123654987. Its first 2 digits (12) are divisible by 2. It's first 5 digits (12365) are divisible by 5. However this is not a solution, since 1236549 is not a multiple of 7.
6. Arrange the Jacks, Queens, Kings and Aces of the four suits { Spades, Hearts, Diamonds, Clubs } in a 4x4 array, in such a manner that: Each row has exactly one card of each rank and one card of each suit. Each column has exactly one card of each rank and one card of each suit. Both major diagonals have exactly one card of each rank and one card of each suit. Euler, the great mathematician, proposed the task of constructing a similar 6x6 array, but instead it was proven to be impossible. Does a 5x5 array exist?
7. ## jelly beans revisited - achieving equality

Good start. Is it always possible to achieve a distribution ratio of 1:2:3?
8. Four towns, A, B, C and D, are located such that their centers form the vertices of a square 1 mile on a side. Town planners want to build a set of roads that connect the four town centers while minimizing the cost, which can be considered to increase linearly with road length. What set of roads minimizes that cost? A B C D
9. Nice work. Bonus solution is the one I had in mind.
10. Bumping this puzzle and offering a clue - unless someone's actively searching for the final case.
11. ## jelly beans revisited - achieving equality

In a previous puzzle @plasmid found that by successively doubling the jelly beans on a plate by transferring from one of the other of three plates, it's possible to empty one of the plates. Suppose the starting number of jelly beans distributed among three plates is a sufficiently nice multiple of 3, namely b = 3x2n. By making successive doubling moves, as in the first puzzle, is it always possible to end up with an equal number ( 2n ) of jelly beans on the three plates?
12. Hi Cygnet, and welcome to the Den. And nice pic.
13. What fraction of triangles in a circle are obtuse?
14. This puzzle is an ancient one that doesn't have a definitive answer so far as I can find. Points given to ThunderCloud and plasmid for proofs of possible answers, but will leave the puzzle open for further comments. I have a criticism of this puzzle, closely related to Pickett's comments, that I haven't seen raised elsewhere: There is no such thing as a random triangle in the plane. How do you pick random points in the plane? We can impose a coordinate system that makes (0, 0) a reference point, and we can add (1, 0) to provides a scale factor and orient the axes, but that's it. What we can't do is pick three arbitrary points in the plane. The origin can be the first point, WOLOG, but the other two, if truly chosen at random, are both points at infinity. It's like asking the average value of the integers. A finite value would, by any measure be disproportionately "close" to the origin. I haven't found any reference to this objection in other discussions of this puzzle. The issue plagues any attempt at a solution. In the analyses plasmid gave us, the two divergent answers are equally correct -- or equally incorrect. They assume one of the sides of the triangle has finite length. But any random line segment in the plane must have infinite length. And if so, then the analysis compares areas that are both of infinite extent. The analysis ThunderCloud gave us includes the premise that "scale doesn't matter" so let b=1, and then let a and c be anything. Same issue: if b is constrained, we can't let a and c be infinite. Or, if b=1, then it's not random over an infinite space, where things cannot be "scaled". Other approaches that I found let the triangle, instead, be randomly chosen in a circle. This preserves angular randomness and permits comparison of side lengths, by eliminating the infinity problem. As a bonus, it gives a unique answer. Follow-on puzzle: What fraction of triangles in a circle are obtuse?
15. In some cases we can get around this point. We could for example divide the plane into increasingly small squares. If we picked a point at random we could say it lies in all squares with equal probability and then count squares. Since a finite circle includes a finite number of squares but excludes an (uncountably) infinite number, the fraction of squares inside the circle is no longer indeterminate -- it's zero. That would allow us to reach the reasonable conclusion say that the probability of hitting a finite circle embedded within an infinite dartboard is unambiguously zero. So in some cases where we're picking points at random we can start out picking very small areas, getting an answer, then take the limit of that answer as the areas go to zero. In cases where we're dealing with two infinite areas, however, this approach does not work. (Unless perhaps if the infinities are of different cardinalities.) Using these "geometric probabilities" is something like saying that points have equal "density" everywhere. It's kind of a reasonable approach, but it's contradicted by the point that you make, namely that there is a surjection between the interior and exterior of a circle.
16. You can fit all the answers on a sheet of paper. Nice thinking tho.
17. Sorry, I should have pointed out this was a thought experiment and all three hands are distinguishable.
18. Those are still not all of them.
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