# Tag Info

## Hot answers tagged construction

34

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

26

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

24

Simply rearranging the symbols used in the intended solution.

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 ...

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:

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

15

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 ...

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.

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.

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. $(\mbox{all games for X}) \... 10 Another solution (with diagonals as bonus): 10 4 6 21 18 7 20 2 28 12 3 5 1 15 14 24 Things that multiply to 5040: each of the four rows each of the four columns each of two diagonals four center cells four corner cells two middle cells of the top row two middle cells of bottom row two middle cells of the rightmost column ... 10 I can do it in just: Initial configuration: First: We have: Now: We get the mark: Finally: You get: And the required distance is: Why this works: 9 Select an arbitrary point. We will call this point O. Set the compass to some radius and draw a circle centered on O. Choose a point on circle O, we will call this point A. Using the radius AO, find the intersections of circle AO with the original circle, call the upper point M, and the lower N. Repeat this process using the same radius and points M and N ... 9 You can fit them together like this (diagram not perfectly to scale): which uses (Highly sophisticated mathematical solution technique: print on paper, cut out the shapes, push them around and see what happens.) 9 9 This solution works: (Places where the wires are close together can be pushed apart a bit - there aren't any crossed wires, though.) My strategy: 8 The best I can do so far is 5. Edit: Got 4! 8 Edited: added a drawing for the first step. Edited again: Dr Xorile very clever solution eliminates the second step. 8 8 I must be missing something because I'm getting a lot more than 54 mates-in-two for the position below? I have 116 listed, although I'm doing this by hand so there may be some errors included. Is there some rule I haven't considered? 8 I think you could make cuts as follows 7 My answer: 7 The knight cannot succeed when$N$is odd. Color the squares in checkerboard fashion, with a corner black. When$N$is odd, the removed square is white, so there will be two more black squares altogether. Since tours must alternate colors, none can exist. We prove by induction that, for even$N$, an$N$-palace (i.e. an$N\$-cube with its center removed) can ...

7

Here is a solution with 9 tetrominos: I made some calculations for an upper limit to see if bruteforcing 8 tetrominos would be possible, and it doesn't look good: 64 starting positions for the first tetromino 60 for the second 56 for the third ... per starting position 2*3*3 possibilities End result: 3.7e23

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