51

One possibility, if starting with an operation automatically prepends Ans: Why does this work?


39

Solved in 10 moves including "Add a comment" and "Add Comment" buttons at the start and end I was muddling around in the "inspect element" for way too long trying to find something I could use in the html.


35

The answer should be Here is how you do it.


27

This is hugely inspired by @rhavelka's answer, so make sure to upvote him/her as well! Just found an alternative of where the phrases can come from:


26

It might be possible to do a little better, but I think I can get Explanation:


24

The trick is this:


21

In general, with logic puzzles it's important to mark not only the parts of the solution, but also things you know can't be part of the solution. Here, that means you should mark segments that you aren't allowed to draw, because there would be no way of completing the puzzle if you did. For example, there can't be a connection between F1 and F2, because F1 ...


20

Not sure if minimal, but here is a way to do it in Method Also, as Jaap Scherphuis points out in the comments


19

Probably not optimal, but here's an approach that saves at least


19

I wonder if this is the answer you have in mind, because it feels kind of like cheating... Also depends on the calculator's syntax, so tell me if it's well-typed. Edit: A technique that works (in five presses!) even if the previous one doesn't: Also, a hint to others regarding "noble" answers:


18

If the person can stay in those 2 minutes, the problem will be unsolvable because he can only move if and only if we check his room. Otherwise, here is a solution with exactly 15 14 (thanks @JaapScherphuis!) checkings, just in time!


16

The same idea as APrough's answer, but presented visually instead of with words: 📅 🏠 🏜️ 🏜️ 🏜️ 🏜️ 🏜️ 🏁 0 🤠🍗🍗🍗🍗🙂🍗🍗🍗🍗🙂🍗🍗🍗🍗 1 🤠🍗🍗🍗🙂🍗🍗🍗🙂🍗🍗🍗 1 🤠🍗🍗🍗🍗🙂🍗🍗🍗🍗🙂🍗 2 🙂 🤠🍗🍗🍗🙂🍗🍗🍗 2 🙂 🤠🍗🍗🍗🍗🙂🍗🍗 3 🙂 🙂🍗 🤠🍗🍗🍗 4 🙂🙂 🤠🍗🍗 5 🙂🙂 🤠🍗 6 🙂🙂 🤠


16

Alternative solution: Then here's how that works Here is the total number of rations and helpers required


15

This is my first answer on the puzzling stack exchange and I'm really not used to explaining things like this so it's probably going to be rather convoluted. I'll probably come back and edit it later when I can figure out how to make this clearer. Solution for a maximum of 4 weightings: First divide the coins into 4 equally sized piles. There are now 3 ...


15

Answer: Strategy: Visualisation: Proof:


13

A complementary result of tehtmi's answer: Using the same strategy, is the maximum number of coins you can get, and +1 is impossible. Proof:


13

The optimal strategy takes Here's why this is necessary: And here's a way to do it: Fun fact:


13

I think this is a classic (and probably a duplicate): To separate the two groups cleanly


12

Not a definite answer but I narrowed it down to 2 possibilities. It's either In this case or the second possible answer is Because


12

This is a huge question, and I think people can help each other here so I'm going to start with some initial strategy: I have set up this board: Which allows for this move: For a score of 2372 (Edit: confirmed by OP)


11

You can solve the problem via integer linear programming as follows. Let $n$ be the number of coins, so we need at most $n$ bags. For $b \in \{1,\dots,n\}$, let nonnegative integer decision variable $x_b$ be the number of coins in bag $b$, with $x_b$ nonincreasing. Let $z$ represent $\max_b \{b\cdot x_b\}$, which is the number of coins the king will take. ...


11

If you are unsure about the puzzle, you are probably misquoting a classical puzzle as Rand Al'Thor mentions. If you cannot distinguish the type of the coins, regardless of how you make two piles, there is always a way to swap a heads-up and a tails-up coin between the two piles, changing the heads-up counts. So you cannot guarantee the counts are the same. ...


10

The answer is Yes.


9

Here is a proof that hexomino's answer of is best possible. Here's how. The alternative is (This can be expressed more concisely using the language of graph theory, but it seemed better not to.)


9

If the first player is the one who says either 2nd January or 1st February, then the winning player is Proof: The first player The strategy is to Note: I assumed that the game only goes over one year. If players can go on to the next January, then


9

EDIT2: My solution was already accepted, but (I assumed) far from optimal. I finally made myself a prime checking algorithm and checked the 32 numbers that allow maximum scoring for the main 'word'. Edit: Score increased to 3532


9

Can we remove the magic? This is the path I took to make a magic square into the grid: First, let's take the magic square given: We need to make it to: Notice So with that, After that, That helps out a ton, as the rest is pretty trivial: Then, Lastly, Well, say goodbye to the magic in the 3 by 3 magic square, and say hello to a normal 3 by 3 square!


8

Assumption: Shifty coin only switches if weighed. I believe this is optimal solution; if (case 1) otherwise (case 2) If it is Case 1; if (Case 1.1 - They are equal) if (Case 1.2 - They are not equal) then if Case 2 happens (no balance at first weighing); if there is balance (Case 2.1), (also meaning it was heavier at first) if no balance again as ...


8

We can limit the loss to Bags are as follows: Examples of results for different sizes the knight could pick: This seems even closer to optimal because Some notes on technique and strategy: I put together a spreadsheet which In the final configuration shown above, it looks like this: Some explanation: The spreadsheet can also automatically calculate ...


8

I think I can do it in How to achieve it: Now, why is this optimal?


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