# Tag Info

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Notation: (P,Q,R,S,T) denotes that form P is in Switch 1, Q in Switch 2, etc. If Switch 1 is wrong, then If Switch 2 is wrong, then If Switch 3 is wrong, then If Switch 4 is wrong, then If Switch 5 is wrong, then Answer

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We see that We then determine So that means that So Sparrow got The answer is therefore

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Two possibilities. There are no Arriban villagers. There is at least one Arriban villager.

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If I can resurrect this one, I think I've spotted the links between all the previous partial solutions, worked out an element that had been missed by all, and can explain the additional hints in the text. Your girlfriend wants you to buy her: Puzzle 1 (solved by @PeregrineRook): Puzzle 2 (solved by @ThirupathiThangavel): Puzzle 3 (worked on by @Techidiot ...

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Initial information, re-parsed We have a sequence of length 2012, in which each term is either N (nicely) or A (always mad) or C (mad converted from nicely). Among terms 2009, 2010, 2011, 2012, three are N and the fourth is not, and that's never happened with any previous set of four consecutive terms. Important deductions If any term is C, then If we ...

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I think, all the other answers (though correct) are way to complicated: of course the right solution is But the reasoning is much easier:

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That particular puzzle is meant to be solved by trial and error. The Super Hint for that puzzle says: "Thankfully, it doesn't matter if you get it wrong and make the porters buckle under the strain." If you input an incorrect configuration, it shows which of the two porters is carrying too much and lets you try again. The solution is not deducible ...

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First, we should determine what the sequence is. Now, does a 6 ever appear in this sequence? Quick detour Back to the show

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Up to symmetries of the board, there aren't very many possible moves for the first player: Does this strategy work?

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I think the answer for $14 \times 14$ is Achieved as follows While the best I've achieved for $9 \times 9$ is Achieved as follows

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Form A is available at Assume

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The missing square should look as follows Reasoning

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This is a more graphical answer with 1's for mad frogs and 0's for nice frogs. Here are the four possibilities for 2009, 2010, 2011 and 2012 No we can go backward bit by bit for every sequences, using the two given rules and the knowledge that an ashamed frog never happened before. We see that there is a recurring pattern for each possibilities. Since 1, ...

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Let $R(n)$ be the number of codes of total width $n$, so the question asks for $R(14)$.

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First observation: Now is optimal because

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Clarification of the Problem and Very Partial Answer Transforming the story into the expression. A through I are distinct digits from 1 to 9, X and Y different digits within 1 to 9, and m and n any integer. AB can be CD can be m and n EFG can then only be Exhaustive list of possible values of EFG:

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The way it is written currently: If the question means up to but not including 10 lbs then:

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building on Omega Krypton's answer: but but...

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This answer is no different than @RShields, but it adds a visual to the answer that might be useful for some.

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Solution: Logic:

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There are at most 1346 inhabitants in Arriban if all of them are at the festival. If not all of them are at the festival, there are 1352 inhabitants.

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For $14$x$14$ I got... With this ... Or with this...

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From the statements So And And Thus So

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this puzzle building upon @Dark Thunder's answer

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Working backwards as in the classic... A,B,C,D,E Two pirates remaining: Three Pirates Remaining: C,D,E Four Pirates: B,C,D,E Five Pirates: A,B,C,D,E

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Using: we find:

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After round one, we know that the robots have one of: After round two, we know: After round three, we know: After round four, we know:

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