BrainDen.com - Brain Teasers

# bonanova

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1. ## Fettered random walk

Here’s a challenge: I think this puzzle can be solved with almost no math at all. Like, think of a single “what if” question to ask that changes the conditions slightly.
2. ## Line segments of two types

Six line segments serve to connect pairwise any four points in the plane, no three of which are collinear. It's clear that no placement of the points permits all six to have the same length. How many unique placements permit the segments to have only two distinct lengths? Example:
3. ## Limit of a shrinking function

I have a strong feeling, and I'm working on a proof, that
4. ## Limit of a shrinking function

Which way should we evaluate this? [ [ [ [ [ x^(x/2) ]^(x/4) ]^(x/8) ]^(x/16) ]^(x/32) ].... or x^[ (x/2)^[ (x/4)^[ (x/8)^[ (x/16)^[ (x/32) .... ] ] ] ] ]
5. ## Random passengers

One approach, among several, is to and
6. ## Squirrel

I guess I've always been a little confused about what is being asked. Sharing is not permitted, yet an "average" result is requested. If average values are used, then two holes is enough. If there is no sharing or averaging, survival is assured only by the (very unlikely) worst case of 100 holes. If the question is what is the expected number of holes that together yields at least 100 nuts, we have an answer from simulation. Is there a way to say precisely what else might be needed?
7. ## 100 white marbles

@harey, some guidance about completing a formula would be appreciated. I'm interested in the "surprise."
8. ## The holy mans walk up the mountain

Yes you cand do that. The elevation of the two men as they ascend and descend are continuous functions of time. If at one time one is greater and at a later time the other is greater there must be a time when they are equal. Since they both stay on the path, and if the the slope of the path never changes sign, then they meet at that time. If the slope does change sign, then multiple points on the path will have the same elevation and there can be multiple times when the elevation of the two men are the same. They meet at one of those times.
9. ## Shortest set of lines in a square, revisited

Good start -- That's shorter than 3, which comes from any three of the square's edges.
10. ## Staying dry in the rain

For the purposes of this puzzle, consider our old friend Albert to have the shape of a rectangular paralellepiped (when he was born the doctor remarked to his mother, I don't explain them, ma'am I just deliver them) just meaning a solid having (six) rectangular sides. At a recent physical exam, Albert was found to be 2 meters tall, 1 meter wide and .20 meters thick (front to back.) He maintains his geometric rectitude by never leaning forward when he walks or runs. So anyway, Albert, alas, has found himself caught in a rainstorm that has 1000 raindrops / cubic meter that are falling at a constant speed of 10 meters / second, and he is 100 meters from his house. Just how fast should Albert run to his house so as to encounter as few raindrops as possible?
11. ## Baggage on a conveyor belt

@plasmid Nice solve. The puzzle itself was more a math exercise than a puzzle -- that part was thinking it through and setting up the calculation. I thought is was interesting in that you can sort of envision the setup and know that it had to be e.g. greater than 1/3, but maybe not as great as 2/3, so the question was would it be greater than 1/2?
12. ## Baggage on a conveyor belt

At a busy airport a conveyor belt stretches from the runway, where all the planes land, to the baggage claim area inside the terminal. At any given time it may contain hundreds of pieces of luggage, placed there at what we may consider to be random time intervals. Each bag has two neighbors, one of which is nearer to it than the other. Each segment of the belt is bounded by two bags, which may or may not be near neighbors (to each other.) On average, what fraction of the conveyor belt is not bounded by near-neighbors? Example: ----- belt segment bounded by near neighbors ===== belt segment not bounded by near neighbors ... --A-----B==========C---D--------E=============F---G-H---I-- ...
13. ## Baggage on a conveyor belt

Yes there is a rule, L'Hopital's rule. Basically you can just replace functions by their derivatives to resolve indeterminate values. Or, you can just evaluate expressions and see how they behave: It's intuitive that the exponential dominates for large x if you think of say x / e3x, instead of x e-3x.
14. ## Squirrel

I guess we can compute expectation value as well:

16. ## Grabbing marbles

You are wearing gloves while trying to retrieve the marbles contained in a bag. Because of the gloves, you're doing a pretty poor job of it. With each grab, you are only able to retrieve a random number of marbles, evenly distributed between 1 and n, the number of marbles currently in the bag. With 30 marbles initially in the bag, how many grabs do you expect it will take to retrieve them all?
17. ## Squirrel

There is survival in numbers.
18. ## Three matches

If we place four matches in the form of a square, they form 4 right angles. If we place them like a hash-tag (#) they form 16 right angles. If someone removes one match, can we still form 12 right angles? (No bending or breaking of the matches is allowed.)
19. ## Grabbing marbles

The cars in front of the slowest car are the remaining cars. (Each slowest car captures those behind it.)
20. ## Grabbing marbles

I think this puzzle is exactly the same as the traffic jam puzzle I posted recently, although that's not obvious at first glance. @plainglazed found a solution in which he formed clusters of cars by recursively locating the slowest of a group of cars, assuming on average it was in the center of the remaining cars. This corresponds to "grabbing" on average one-half of the remaining marbles from the bag. Picture the marbles in a line and, grabbing a random percentage of them starting from one end. This has to end up having the same number of marble grabs and car clusters. It leads to a logarithmic answer, but to the wrong base -- it should be the natural loge, not log2, which gives too large an answer. Instead of decreasing the number by 1/2 each grab, the remaining number is decreased by 1/e each grab. Same must go for locating the slowest of the remaining cars. @plasmid found a solution that leads to Sum { 1/k }, which as you point out is ln { n } + gamma, and is here confirmed by simulation. What I can't find is a corresponding analysis for the marbles problem (although plainglazed's approach is applicable to both problems) that will lead to that same sum, instead of yours, where both sums appear to be correct. This has been fun to think about. .
21. ## Take a guess

If you chose to answer this question completely at random, what is the probability you will be correct? 25% 50% 0% 25%
22. ## Grabbing marbles

That agrees with simulations I ran, which gave, approximately, Nice solve.
23. ## Staying dry in the rain

You're right. And it's not much of a puzzle after all. You have to allow Albert to lean forward as he runs to make this puzzle at all interesting. But even then, the obvious solution is to have Albert lie horizontally and crawl at infinite speed. Only the top of his head gets wet then. I ran into this puzzle a few years back and "solved" it, more interestingly but also more incorrectly, by multiplying his front and top areas respectively by sin theta and cos theta where theta was determined by his speed compared to the speed of the rain, and some other stuff. It was nonsense.
24. ## Dividend please (on steroids)

More clues...
25. ## Dividend please (on steroids)

Here are the placeholders for a long division, solvable, even with none of the digits filled in. The quotient has been placed to the side. It has a decimal point, not shown, and its last nine digits are repeating. Meaning, of course, the last row of X's replicates a previous row. Can you piece together the dividend? -------------- _________________ x x x x x x / x x x x x x x ( x x x x x x x x x x x x x x x x x x ----------- x x x x x x x x x x x x x ----------- x x x x x x x x x x x x x x ------------- x x x x x x x x x x x x x ------------- x x x x x x x x x x x x x x ------------- x x x x x x x x x x x x x x ------------- x x x x x x x x x x x x x x ------------- x x x x x x x x x x x x x x ------------- x x x x x x x x x x x x ----------- x x x x x x
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