# Tag Info

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I have a hunch that the answer is Explanation: Continuing this way, we see that

28

Here is a simple strategy of how they could do it Proof

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Using only combinations of either single or double button presses: Using 1 press Using 2 presses Using 3 presses Using 4 presses Using 5 presses Total For anyone curious as to my internal process of coming up with these values, I made a quick chart of each type of press combo and how many variations there were for each type, with x indicating it has ...

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First of all, It will probably make a difference So let's concentrate on the other ones We can already stop here; so

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4-button method

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This is, I'm sure, answered somewhere else. It is also related to Pascal's triangle. Simply fill out the grid as follows: In this grid, each number represents the number of ways of getting to that particular intersection. And that number is precisely the number of ways to get to the intersection below it added to the number of ways to get to the ...

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Perform tests of nine sheep on all but one sheep according to the illustrated patterns: The two important properties exhibited are The claim is that given a set of test results there is at most one possible group of five wolves. Suppose instead that some set of test results could have been produced by two different groups of five wolves A and B. Then both ...

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Answer: Consider for instance You get the sequence of numbers In both cases,

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You have all the possibilities using only single button press Then the possible combinations using one pair are: Then you have the possibilities using multiples pairs: Adding all cases, you get: Which is the expected answer from the question!

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To specify a rectangle

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I think the correct password is Reasoning

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There are more possibilities than one expects because Here is the full list, after removing symmetries; I have marked some of the more surprising ones: So there are A visual representation:

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Thinking out loud, not a solution yet, but spoilery enough that I didn't want to put it in a comment: However, Still-not-an-answer UPDATE: However, I also notice that the situation is not symmetrical: we may know a way to find $k$ wolves among $n$ sheep using $t$ tests, but that won't help us at all to find $n-k$ wolves among $n$ sheep. (Under the ...

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Here's another solution. Unlike the highest-voted answer, it's adaptive. However, it may be easier to put into practice. Then the apprentice does the following: And the master does the following:

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Of course @Gareth_McCaughan got this well-known puzzle immediately. But for people who aren't up on their combinatorics, here's the same calculation in a way that seems easier (at least to me). There are 9x9 = 81 corners. For each of these there are 8x8 = 64 corners that are not in the same row or column. Each pair of these makes a rectangle. But then each ...

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The cards will again be all face down . . .

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It takes Suppose Now

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Proof:

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Here is a method for constructing a solution without the use of a computer. [EDIT: This generalises to any even number of sets, but not to odd numbers. See the edit at the end for a different method that I think will work for any number of sets.] EDIT: Here is a different method that I believe generalises to any number of sets. I will use 5 sets in this ...

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Well, let's see. It's possible for 1 to 5 buttons to be pushed to create a combination, so let's calculate the number of possibilities for each individually, assuming you can pick the order of the buttons to be pushed. 1 Button 2 Buttons 3 Buttons 4 Buttons 5 Buttons Total

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$n=2^k$ button method Explanation:

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Continuing from Arnaud Mortier's observations, Using that knowledge and a bit of trial and error,

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We can Now the question is This means Of course, Further requirements of the strategy To be able Going We can continue The big mess

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Fact 1: Fact 2: Proof: Exhaustion... Maximizing possibilities: Notation: (x,y,z) : highest possible value Possibility 1: (1,2,?) Possibility 2: (1,3,?) Possibility 3: (1,4,?) Overall:

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