New answers tagged

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Thanks to boboquack and Gareth McCaughan for completing the proof.


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You're not going to like this, but: "OK, machine, listen up. I'm going to describe an algorithm for you, and it's a little bit complicated. First of all, let me define a Turing machine for you, along with a simple numerical representation for what's called a Turing machine's state table. [Do this. I promise I can.] And now let me define for you the ...


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The stock Woman A: 30 apples (originally sold at two for a penny) Woman B: 30 apples (originally sold at three for a penny) Indeed, if each sold their apples individually, there be a total income of 25 cents (Woman A = 30/2 = 15 cents; Woman B = 30/3 = 10 cents) It would seem that this is the same result you get by combining the stock and selling five for ...


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Picture-driven Explanation Answer: Explanation:


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Let us denote the ages of Person 1, Person 2, Person 3 by $x,y,z$ respectively. We'll assume that $x,y,z$ are positive throughout. The product of the 1st person's and the 2nd person's ages is $311 \frac{2}{3}$ plus the 3rd person's age. The sum of the 1st person's age and the quotient of the 3rd person's and the 2nd person's ages is $41 \frac{17}{24}$ ...


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I don't think there's anything super-clever you can do. The important thing is to be systematic so you know you aren't missing any possibilities. The overall approach I would take is a "depth-first search". Draw the network of vertices and edges on a piece of paper. Leave out H which you aren't allowed to use. Give names to all the vertices -- ...


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I wrote a program to do this and it gave me the answer of 11 static void Test() { int[][] numbers = new int[8][]; //numbers[1] will contain all of the points connected to point 1 //numbers[2] will contain all of the points connected to point 2 //and so on... numbers[1] = new int[]{2, 3, 4}; // A numbers[2]...


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Looks like the apparent relation between the shapes of the arrow heads and the operation they imply is a red herring. And even if they aren't there is no reliable clue as to what open versus filled heads may signify.


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Occam's razor says the simplest explanation is the best.


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4

It's simply just So Therefore the missing number is


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Here is an easy solution if we assume that every number goes to $4$. Notation: Let $L(n)$ be the number of letters in our number $n$. Let $S(n)$ be the number of steps it takes to get to $4$, (so $S(n)$ gives us the $x$ number in the question). Claim: There is no finite upperbound on $S(n)$. Proof: Suppose for sake of contradiction that there are no chains ...


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An easier proof, starting from jafe's idea: Let $d(n)$ denote the number of letters used in writing down a number $n$. Let $S_0=\{4\}$, and recursively let $S_{n+1}$ be the set of natural numbers $k$ for which $d(k)\in S_n$. (Equivalently, $S_n$ is the set of numbers such that repeating the step $n$ times reaches $4$.) Claim: $S_n$ is finite for all $n\geq 0$...


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I claim that $x$ is unbounded. Let $L$ be the function that counts the number of letters in our number. Consider centillion $=10^{303}$, just as million-million is $10^{6+6}$, centillion-centillion is $10^{303+303}$. For brevity we notate the n-fold concatenation of centillion by (centillion)$^n$. Note that (centillion)$^n = 10^{n\cdot 303}$. More ...


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Solution


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[Revised to cover both parts of the puzzle, beginning with the original post.] While the puzzle has already been completely solved by Jaap Scherphuis a geometric approach works very nicely for the puzzle’s first part that asks for the width of the river. As time flows downward (as well as downstream if the ferries are carried by a current, thanks to Pspl’s ...


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Here is a simple argument to deduce the solution to the first question without much arithmetic. The second question is not quite as nice, but now we know the width of the river and the relative speed of the boats, so it can all be calculated fairly easily.


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It's not clear what you mean by "a 10×10 binary puzzle". But it's possible that the designers of the puzzle accidentally left the puzzle with multiple solutions. A pure logic puzzle shouldn't have multiple solutions, but authors do make mistakes.


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Completely new and improved answer


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The answer is indeed: This is because, simply:


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Here is an explanation for 18 as a solution:


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Maybe the answer can be Because Of course


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My response: Explanation: And from the higher ground we see:


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I have a theory: I think that the original puzzle looked like that: � is the unicode replacement character, which was shown because the original symbols (some emoji?) could not be displayed (see Mojibake). So, what were the original symbols? We don't know, but we can try to make the puzzle solvable by substituting them: Which can be answered trivially. ...


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This is probably a bit of a stretch, but maybe Reasoning: Also:


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Its pretty sure that, If the camel takes all the 1000 bananas at once to the end point,he would be left with 0 bananas at the end. So, there needs to some Intermediate Points that are in between O km and 1000 km. So, to prove my point, Let us assume 4 points. let's called point A------>Beginning Point point B------> End Point point Q------>First ...


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