New answers tagged

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It is not possible with any finite or countably infinite set. Take any point P in the set. For each other point Q in the set, draw a line between P and Q. Then take the angle between line PQ and the horizontal axis. This forms a set of angles A. If S is countable, then the set of angles A must also be countable (since there is at most once angle per point ...


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Perhaps the country is The numerator is clearly It happens that So the "fraction" Now, what else do we have? The best I have been able to do at combining these things to make a country is as follows: I am by no means 100% convinced that this is the intended answer: But The actual homeland


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I believe the set of points (x,y) such that $x y = ±1$ satisfies the conditions. The set consists of 4 segments of hyperbolas. Any straight line crosses at least 2 of these segments resulting in 2 to 4 intersections. Except for the lines x=0 or y=0 which cross none.


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The smallest currency denomination is 1 cent. Percentage needs to be integer. Answer:


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Going by the information regarding currency denominations available here, I think the answer is: Possible strategy 1: Possible strategy 2: Possible strategy 3: Why this is optimal:


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Here is my intuition on it, which is not mathematically rigorous. Answer Explanation As stated by Hint 2, running between 2 columns of zombies can't be done infinitly as zombies get too close to us after some distance. Yet it seems a good starting point. Let's consider we get to point 50, 50 and then run upwards (between the 2 columns of zombies). We will ...


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The OIES entry for the sequence A047837 calls this "Honaker's triangle problem" and says that the optimal $N$ is conjectured to equal: $$ a(n) = \max_{r=1\ldots n-1}\left\lceil\frac{T(T(n)-T(r-1))}{n-r+1}\right\rceil $$ (sequence A047873), where $T(n)=n(n+1)/2$, the $n$-th triangular number. The OIES entry refers to the book The Zen of Magic ...


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Okay.... I googled "A game that doesn't exist" and wound up with "Polybius (square)" which also came up in the commentary below. Hadn't heard of that before. A game that has existed for over 1500 years" presumably refers to Chess. I thought that might be the 'key' for a simple version of the square or an 8x8 square but it didn't work,...


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For n=100, is the answer? How? For other n How are we sure it's a tree?


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I'm not sure about the bonus but here's the best you can do for the main part Proof that this is the best Lower bound for the bonus Improving this bound for larger $n$ (credit to Greg Martin in the comments for this idea)


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I expect there are many. e.g.


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Here is one example


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A general, inductive approach for finding the limit $f_E(N)$ of $N$ trials with $E$ eggs: EDIT: In fact, the inductive definition lends itself to a closed-form expression, with a slight change of perspective:


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After recalculating the table I believe that there is a winning strategy for A, as follows: A chooses 1 first and ends up winning by 1 point. See Tables below showing the way an optimal game could progress. Spoiler space added as I cant work out how to hide a table. An optimal game would progress as follows (Asc and Bsc are the scores for A and B after ...


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60 / 2(10+5) = 60/2(15) = 30(15) = 450


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I'd like to complement the other answers by a more formal explanation.


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A score of 41. Or a very abstract score of 140.


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The upper bound for $K_3(13,3)$ here yields This guarantees $10$ correct out of the first $13$ questions, so we can surely do better by considering all $15$ questions. A lower bound is the sphere covering bound $$\left\lceil \frac{3^{15}}{\sum_{i=0}^5 \binom{15}{i}2^i}\right\rceil = 118.$$


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We can slightly improve on 65.


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Here is my explanation. He buys the goat for 60. He sells it for [sealed box of money who knows how much is in there.] Then he buys it again for [sealed box of money who knows how much is in there] plus ten. He has spent a total of 70. He sells it for 90 so his profit is 20. This doesn't rely on adding two separate profits and then wondering if one of them ...


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The answer is and here's why:


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It can be done using Outline: Detailed example showing a few rerfinements:


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The logic error here is that coming up with a profit of only 10 subtly double-counts the two middle steps. As others have pointed out, the basic math is $-60+70-80+90=20$. The faulty logic is 10 profit, 10 loss, and 10 profit summing to 10 profit overall. However, the first 10 profit comes from $-60+70$, the loss comes from $70-80$, and the second profit ...


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The absolute simplest way to prove this to your friend is to offer to actually perform these transactions with you - and to make it worth his time, you'll pay him $15 to do so. Ask him to sell you something (a goat, or any suitable substitute) for \$60, then buy it back for \$70, then sell it to you for \$80, then buy it back for \$90. According to him, he'...


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If the leading digit could be zero, which it can't, but if it could, then by This results in Note that @obscurans pointed this out in a comment


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Other Solutions: Here are a few ways to get more than 8:


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It is Proof:


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The LHS is than the RHS. Here's how we're going to prove it. OK, let's get started. We want to know whether $\frac23\left(\sqrt5-1\right)^3<\sqrt[3]2$. So, That's the first step completed. Now Nearly there. Finally, (The above is fairly long, but only because I've gone into quite a lot of detail. The actual calculation is rather quick, and for those ...


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With N teams there are Edited to improve solution:


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Let's focus on the A few observations: Conversely: This already resolves C as To avoid tedious edge cases I'll assume that the enclosure is a rectangle with one side completely open. B A


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So it is


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Note that So, let's work backwards: For the record:


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Well, mean is infinite (easy enough to see): Median is finite though. In code I slightly simplified the problem - function [n,m,k] = so_coins(n) % n=number of coins left in the urn. m=0; % number of coins taken so far k=1; % steps performed - end condition. while (n > 0 && k < 1000000) % stop after 1m steps of taking coins out. r=rand(n,1)...


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This answer draws freely from other answers, in particular, @Gareth's and @Steve's. As pointed out by others our strategy Let's assume that the billionaire gets cold feet once there are a total of M (M>100) coins in the pot. We can give a closed form solution for the expected time $E_{100,M}$ until either this happens or the game ends regularly. This ...


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What a nice coincidence when I found StackExchange's featured hot topic to be almost the same question that I asked some days ago at CrossValidate SE: Intended selection bias. Here I describe the following procedure: Consider a bag of balls of two equally distributed colors. If I pick a ball uniformly at random it is a red or a blue ball with equal ...


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Between 500K and 500M gold coins, depending on the exact net worth of the billionaire As others have explained, on average one gold coin is replaced for every coin you take. So the expectation is that the game goes on as long as the billionaire can continue replacing the coins. A one-ounce gold coin sells for around 2000 US dollars according to this website ...


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With a bit of lateral thinking, this puzzle does have a solution (several, really): First off, use @A_username’s insight to realize that the numbers cannot be in a base in which the parity of the 13 is different from the parity of the 8 and 6. Odd number bases would allow it to have the same parity, so we need to assume the numbers are in such a base. ...


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Assuming adjacent means they share an edge, then a trivial solution where some white squares exist is If adjacency includes squares that touch diagonally, then something simple like these should work:


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Here's a solution to both problems


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This works because


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Score: 98 100! The O tetromino appears four times, all others appear three times. Sadly, there's no symmetry to exploit like there was in my answer to the trimino version of this question. The grid:


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Short answer: Long answer: The solution to the problem:


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I agree with the answer some others have given but I don't think their analysis is correct. First of all, So, let the expected number of coins you get, when the urn starts off with N coins in it, be f(N). We want f(100). On each turn, the billionaire chooses a number m from a geometric distribution with parameter 1/2 (note: there are two different things &...


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Attempt at a bit more rigor in an answer. Consider first the strategy of taking 1 coin each turn: For other strategies: Comments:


2

Let's assume the integers are $x,y,z$ and $-w$. All four equatins summed up gives $2x+2y+2z+2w=8+6+13+8$ or $2(x+y+z+w)=35$. Left side is even, right side is odd, so no integer solution exists. (see correction below from Jaap Scherphuis to $2x+2y+2z+w-w=8+6+13+8$)


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What is the maximum and minimum expected number of coins you can extract from the game?


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