50

I think this arrangement works for the bonus question:


46

I think this will do it


36

I don't know how to do the formatting (thanks McMagister for the edit) but the answer is


27

Ernie's jigsaw puzzle isn't as straightforward as it seems, as it's actually: One way of assembling the pieces legally is: How will you know when you have succeeded? PS Ernie definitely has a sense of humour about him. After all, when you texted him with "Bored", he replied...


26

This seems to work: And the position looks like this: Apart from the symmetrical solution, this might very well be unique:


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

Simply rearranging the symbols used in the intended solution.


24

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


22

A possible solution is: 10 8 3 21 12 15 14 2 1 7 30 24 42 6 4 5 Strategy $$5040=2^4 \times 3^2 \times 5 \times 7$$ First I decided where to put the multiples of $7$ and $5$. Then I multiplied proper exponents of $2$ and $3$ to each cell. I started with: 5 1 1 7 1 5 7 1 1 7 5 1 7 1 1 5 This formation ensures that the number of ...


21

Vepir has helped twice in this answer, first in spotting a mistake and then with an improvement in the number of moves. Please got upvote their answer too if you like this one. Here is a position with This position can easily be reached in


20

Here's my answer: From this point on, there is no spoiler text, because it makes it easier to format, at least for me. T E N I N E F I O N E V O G X R V I U H H I E F R T W O S Z I started by figuring out the letter density in the overall puzzle. The letters break down as follows: E 10 N 5 I 4 O 4 T 4 R 3 F 2 H 2 S 2 V 2 G 1 U 1 W ...


20

The 'hardest' possible Irregular Sudoku has and it looks like this:


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,


17

Here is yet another solution with 9 pieces. This one is nice and symmetrical. I have been trying to think of a way to show that 8 will not work by arguing in terms of the number of edge squares that need to be covered. However, I have not got very far with this. Here is a diagram I have been pondering. Update I haven't really given this a lot more thought....


17

Probably fails the no letter criterion. Or using Lagrange notation as a workaround (thanks to McMagister) we can also write


16

My answer is Explanation:


14

I came up with this:


14

As far as I know, the only way to figure this out is by letting a computer run through all the possibilities. It is a small puzzle, so this does not take long. First I will assume that you want the final solved position to have the blank in the bottom right corner, with the tiles in numerical order: 123 456 78. (See further below for the results with the ...


14

Score: Position: Moves: Fun fact,


13

Here are my first idea (both sides are essentially the same answer, so the hint fits too): Both positions seem to be independently reachable by a legal game. It might be possible to find a legal game leading to the whole position too, but that would take a bit of time. Before that, I'm going to double check for any simpler solutions. :-) Since OP ...


13

Suppose you can set your pair of compasses to length 1. Then Then we know that the length of the line that would join those two points is The simplest construction would use:


13

Of course we need to use Pythagoras. This leads to the following solution: Here is another more compact solution.


13

This should do it: White is in zugzwang, so black has


12

My position has a fourteenth check too, but that's probably ok. :-) These are the moves: And here's the whole solution uploaded to Lichess. EDIT: Looks like 15 is also doable: The final check feels like such a waste of a piece though; the first fourteen checks are possible with white having only two pieces and a king: Again, here's the Lichess link.


12

I can do Could we do better?


11

Solution: Explanation: I noticed that there are only ${9\choose5 }= 126$ ways for the $X$ to play a game. I noticed that only $28$ of these would not represent a win for $X$ I then noticed that of the $28$ some may be a win for $O$. It turns out that after removing those which would be a win for $O$ there are $16$ left. i.e. $(\mbox{all games for X}) \...


11

A bit of digging around finds a position, taken from the 7-piece endgame databases, that's claimed as a mate in 549 moves. However, this isn't a 'constructed' position; there's no theme to the mating sequence, and the moves are the classic 'no rhyme or reason' dance that shows up so much in these multi-piece endgames. As far as actual chess problems go, ...


11

I can do and this is clearly optimal, because Picture of solution.


10

Another solution (with diagonals as bonus): 10 4 6 21 18 7 20 2 28 12 3 5 1 15 14 24 Things that multiply to 5040: each of the four rows each of the four columns each of two diagonals four center cells four corner cells two middle cells of the top row two middle cells of bottom row two middle cells of the rightmost column ...


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