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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 ...


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

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 ...


15

My answer is Explanation:


12

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:


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

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 ...


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.)


8

The best I can do so far is 5. Edit: Got 4!


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?


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


7

I believe this is possible for Here are the gadgets I need to prove it. Note that it doesn't matter which corner of the rectangle you start since you can always rotate the picture. By using gadgets (1) and (2), you can essentially subtract 3 from the height of the grid. (1) works for grids of even width, and (2) works for grids of odd width (just add ...


7

This problem was edited to be solvable with one solution. However, there may be references in the discussion to my old answer (see edits) when there weren't the constraints 100 total employees or Contractor wage = $8500. We can now easily calculate the salaries from your information: We also now have two other equations, labeling $A$, $B$, $C$ as the ...


7

Edited: added a drawing for the first step. Edited again: Dr Xorile very clever solution eliminates the second step.


6

I fully copied below answer from https://math.stackexchange.com/questions/227285/constructing-the-midpoint-of-a-segment-by-compass Some Googling revealed the following comments to this answer: I know it is possible, but is there an easy way to divide a segment in half with only a compass? – robjohn♦ May 20 at 3:46 I don't know if that's "easy", but ...


6

This seems like a straightforward way of turning the ladder -- the two marked white extra pieces force black to steer the ladder upwards and left, and the two marked black pieces are to maintain number of liberties. Here's the starting position for this solution:


6

Since the keys are arranged in a circular fashion, I think I need to identify both an origin and a direction, no matter which way the key ring is rotated, and no matter how many keys are moved from one side to the other along the ring. This should do it with minimum trouble: You can always return the keyring to this position, and thereby identify each key. ...


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