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Triangle case: If the triangle has three weights, we can subtract the smallest value from each side. That reduces to the case of only two unknown weights and a zero in the third pan. The original problem defines the maximum weight as 40. Rather than examine this exact case, I assume an arbitrarily large MAX weight and solved for that. This will give an ...

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No frills...

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And now for something completely different:

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Although many people have given this same general idea, it seems like they have all over-complicated a very simple solution. Assuming the other members of the circle are cooperative, and pass along messages accurately: The instructions can be passed quite openly around the circle assuming the other members are cooperative, honest and accurate.

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Try this

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The most simple answer would appear to be Or this, for those who think all three numbers need to be used

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This can be :)

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I will speculate here a bit. Given that everybody passes exactly what he/she got and does not change anything.

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The most simple answer (but it can be unsuitable):

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In Excel: or as Word equation:

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Based on your question change, I would look here on Wikipedoa, https://en.m.wikipedia.org/wiki/Mathematical_chess_problem#Domination_problems It shows diagrams for the smallest amount of each piece, minus the pawn, it takes to cover a chess board. The maximum is uselss-the answer is always 64.

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Making an assumption that the intent is to find the number of different ways that 3 rooks could be placed on a chess board without attacking one another. In other words, not the "maximum number", which implies trying to find the right 3 pieces, but simply the number of different ways. Brute forcing it: The first piece has For each of those options, it ...

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I have found this solution

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Could this be @Oray found another one, which might possibly be

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I have a lateral-thinking solution: Not a bad probability!

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I assume that the prisoners decide on a strategy in the presence of the warden, and that they cannot communicate after this. (So the prisoners cannot have any common secrets that the warden does not know about.) Now the warden can distribute the names over the boxes using their knowledge of the strategy. Under these assumptions the optimal strategy of the ...

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I will begin to start answering my own question. It is not too late to win the bounty. Once it expires I will fully explain my puzzle. Hint 1 Explained: Hint 2 Explained: Hint 3 Explained: How to solve the puzzle Please feel free to ask questions if you need help completing the puzzle. If it is possible to extend bounties, I may do this. Bonus Hint ...

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Perhaps I’m misunderstanding yesterday’s edit to the question, but I believe that I can get the score for 2 down to 4 points:

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Another lateral thinking solution Explanations

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(Note: this is more a mathematical problem rather than a puzzle.) As a first approximation, the algorithm can be seen as follows (it can be not optimal though):

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Here are some answers I found. The trick is Just wanted to point out that trick.

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My answer (after the no-computer tag was removed) is

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So far (partial answer): Edit: Tried to complete the answer, but step 8 may be incorrect.

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I'm adding this because I like it and I remember figuring it out as a child. It's nice because the order of operations work left to right and I guess is technically the shortest without the concatenation or modulo functions.

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Disclaimer: I do not own any of the answers. This is only an arrangement of all the answers. Credits to Andrew Smith, Ismael Miguel, Mwr247, Ian MacDonald and Engineer Toast. 2 operators: 3 operators: 4 operators: 7 operators: So far, answerers need to figure out a solution with 2 operators only to update the record.

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So maybe you have to change the separation. I think this is one sample that works: 1v2 3v4 5v6 7v8 9v10 11v12 1v3 2v4 5v7 6v8 9v11 10v12 1v4 2v5 3v6 7v10 8v11 9v12 1v5 2v9 3v10 4v11 6v7 8v12 1v6 2v7 3v8 4v9 5v12 10v11 1v7 2v8 3v9 4v10 5v11 6v12

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Edit: okay, wow, I underestimated this problem! This is really tricky. I'm ran a search on my computer and found the below solution. I'd be interested to know if there's a simpler way to think about this. The search is to find 6 rows of 12 elements where Each row has two 0s, two 1s, ..., Two 5s. Each column is a permutation of 0,1,2,3,4,5. If elements i ...

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I don't know if this is strictly a puzzle but more of a computational programming exercise. Many people on this site will already be familiar with the iterative function in the set-up to this problem as that from the famous Collatz conjecture. As you would expect, much research has been conducted into the properties of this function and there is even an ...

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Starting with 10 points for 0... ...and 6 points for 1 ... and 7 points for 2 ... and 3 points for 3 ... and 3 points for 4 ... and 5 points for 5 ... and 5 points for 6 (or 6 points if double factorial is not acceptable). ... and 5 points for 7 ... and 6 points for 8 ... and 4 points for 9 (thanks to Jaap Scherphuis) So total points is 54 points....

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I’ve made a (possible) discovery about the arrangements of the consonants: I hope the explanation was clear enough - if not, let me know what I can do to make it clearer.

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So far: Edit (continued): Edit 2: thanks to @PiIsNot3 observation about double letters.

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Here's a formula which I believe contains each number once. Multiplication isn't converted to division for readability's sake. $\sqrt{3}$ = $$1 *\frac{3+5}{2+4}* \prod_{n=0}^\infty \frac{\Big[\prod_{k=1}^{13}(6(13n+k)+2)(6(13n+k)+4)\Big]^a}{ \Big[\prod_{k=1}^{13}(6(13n+k)+3)\Big]^b \prod_{k=1}^{13}(6(13n+k)+1)(6(13n+k)+5)}$$ where  a=\frac{6(13n)+6(...

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It starts like Just

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Probably not the answer but the idea had some appeal for me

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Starting with the leftmost card, number them consecutively from 1 to 20. If the sequence fails to terminate then there is at least one card with number m say which changes its state infinitely many times. But then there must a card with number n< m which also changes its state infinitely many times. But as card 1 only changes its state once, this process ...

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I think that it is because Ok, ok. I know that it is wrong, because

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With 2 mathematical operations:

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Answer 1 Suggestion for the minimum Proof that this is the minimum

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I think he went to Reasoning

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A lower bound is Represent the game as a bipartite graph $G$ on the set of cards and the set of positions. The game starts with a complete bipartite graph. Every turn, Bob guesses a perfect matching (of the complete bipartite graph). For each guessed edge $ab$, if Alice says it is wrong, then $ab$ is removed; otherwise, all other edges adjacent to $a$ and $... 9 I was able to find one of the solutions using "paper and pencil". I stopped searching for more solutions after that. It is certainly very very time consuming. In my explanation I name rows as A,B,C,D,E - columns as 1, 2, 3, 4, 5. My search strategy is based on an observation that As you can see this gave me only 4 possible values for$B1\$ Page 2 ...

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[EDITED to add: When I wrote this, the original statement of the problem had an error in it. That's now been fixed and I've adjusted my answer accordingly.] The most interesting of the questions originally asked is the third: is it possible to get from any number to any other number by this process? The answer is First, the easy bit. Now the not-so-easy ...

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The way I would approach this is as follows: Sort the 16 numbers Find sets of numbers that can fit into rows 1, 2, 4, 5. There are 105 possibilities to try for each (in reality a lot fewer, as for each 1st number pick you can eliminate some 2nd number picks, i.e. if you pick 15, the 2nd number has to be > 200-27-36-15-72 which leaves only 4 options, etc.). ...

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This might be a bit "lateral" answer, and maybe one with assumptions, but: Having no answer that satisfies the conditions at this site after 17 hours have passed means it is in no way a fair assignment for any non-insanely-genius sixth grader. The "this site" is an important part; you will regularly see insanely genius answers around here.

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These are the two solutions I found by 80x86 PC: Edit: about pencil / paper / calculator

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