50

I think this arrangement works for the bonus question:


47

@hexomino's answer is correct and well-reasoned, as always. Here's another approach, which to me feels much more.. "axe-to-the-head" is what I'd call it in my native language, so I thought it might be interesting enough to warrant posting. Lower bound: (This is what @trolley813's encrypted comment is saying.) Upper bound:


46

I think this will do it


29

First solution - 50 moves Second solution - 59 moves Current solution - 66 moves (beaten by Retudin - 70 moves and Rewan Demontay - 139 moves) Moves:


26

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


25

You're


25

Since we are talking about a standard game of chess (although with both players co-operating), we know that there are four pieces that cannot possibly make a capture in the series: the two bishops on the wrong coloured squares one of the kings (the other can be the last to capture) the first piece that gets captured (cannot be any of the above). ...


24

The rules of the question state that: On your final grid, a letter (actually several is also mathematically possible) will be more frequent than all the other letters. Your aim is that this letter appears as much as possible. If we interpret this to mean that all letters appearing with the greatest frequency contribute to the score then it is possible to ...


24

Eeny meeny myny moo (or however you want to spell that)


21

Here's an optimal procedure that lets you find out the status of every person after exactly questions: Reasoning: Proof of optimality:


20

I used integer linear programming to minimize the number of unattacked squares. Here is one optimal solution (unique up to symmetry), with


19

Glorfindel's answer is sufficient for the main question. To answer the bonus question: Here is an example: To construct this example, As for a starting position,


18

Suppose you have $r$ red lamps and $b$ blue lamps. What order should they be put in to maximise the score? Now that we know the arrangement, how many red lamps is optimal?


17

Five $8$s


16

Here's a solution with


16

Reasoning: More reasoning as to the number of queens:


15

Jaap (and many others) already solved the problem by calculus (and other kinds of maths), but this geometry based solution had such a nice symmetry to it that I wanted to post it anyway. First, let's start by figuring out the general colour pattern. Given a single blue and N reds, where should we put the blue? Let's put a reds before the blue, and b reds ...


15

I can do it in This is optimal because: More detailed argument for symmetry:


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

I'm pretty sure hexomino's ingenious answer is not what the book intends, and that the mysterious thing is actually a square root sign on a Roman "I". (Which is pretty stupid, because of course no one ever used square root signs with Roman numerals.)


14

Here is a solution with pieces. The colouring of the empty squares ($2\times2$ in white, $4\times4$ in yellow, $6\times6$ in orange, $8\times8$ in red) shows how the pattern generalises to any $2k\times2k$ board. For $8\times 8$ this solution is optimal.


14

The smallest area I can find: The largest area I can find (another edit):


14

The small one, the "double headed snake" Old attempt: the "circle" New attempt, using the same diagonals as Weather Vane and shuffling around 9 of the sides: some results for the problem in general 1. For what values of $n$ do these n-gons exist? Proof: Regarding diagonals: But that doesn't say anything about whether the allowed ...


13

You can get at most This is possible to achieve: So now we've reduced the problem to the variant where Here's an optimal strategy: This is optimal because:


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:


12

Here is a way to do it with four digits Here is a way to do it with three digits


12

I can do Could we do better?


12

An easy upper bound is because each day contributes $8-2=6$ triples out of $8\cdot 7 \cdot 6$. Here's an optimal solution with I used integer linear programming as follows. For each of the $8!=40320$ permutations $p \in P$, let binary decision variable $x_p$ indicate whether that permutation appears. For each of the $8\cdot 7\cdot 6=336$ triples $t\in T$,...


12

Suppose there are $n\geq 3$ chips. It's easy to see that $n$ moves is possible, making each move exactly once (this flips each chip three times). Since making the same move twice cancels out, any optimal solution also involves making each move at most once. Since each chip needs to be flipped an odd number of times, in any optimal solution each chip is ...


12

I think the answer is Reasoning As trolley813 mentions in the comments


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