12

Here is a picture of the 5 graph: A flat version (perodic boundaries) is easier to digest and reveals the fundamental symmetries: and of the 6 graph: and the 7 graph: Flat version (perodic boundaries). We can see that this construction maxes out at 7. Please note how simple and clear this representation is. It gives a good intuition of how the torus ...


12



10

Here's my first go: Method: After a lot of fiddling around, it seemed like a good idea to crowd the 1 and 2 as much as possible. I started with jamming as much stuff around the 1 as possible, while keeping an eye out for a nice spot for the 2. After a couple of tries, this pattern produced the sweetest spot for the 2 at the last possible moment, and the ...


10

Here is a proof that the answer by @Avi is the largest possible. We have the following lemma, which is intuitively clear and also can be proved rigorously: With the lemma, we now see that and adding them together gives the maximum number.


10



10

Start with this 'snowflake' pattern: So, we can do it like this: There's another way to look at it that's based off:


9

Since the surface of a donut (or toroid) is topologically equivalent to a rectangular space with both pairs of edges wrapping, we can represent a toroidal embedding of a graph as such (which is also easier to see than a drawing on an actual toroid). $K_5$ is easy: $K_6$ is a little bit harder: $K_7$ requires serious thought, because I ended up drawing the ...


8

You need First, an important thing to know: If S≤L: If S≥L: Time for some math! The setup: The Important Thing: So, what does this tell us?


8

There is not a lot to go on, but it could be that the price per donut is The two who came in the day before could have bought:


7

My program agrees on the optimal answer! Here are all 4 unique ways of making it, if (big if) I didn’t write in any bugs: The last one is @RobPratt’s answer, rotated and flipped. They all differ only in the position of the last value. Some facts about picking the starting points: The code isn’t particularly elegant or optimized, but the approach lets it ...


6

First try: Approach:


6

Just putting this answer to show my code and explain how it is done; Everything starts with 4 since 1,2 and 3 already placed, so the only possible way to do that, Therefore, I put two square distance between those two ($1$ and $3$) at most in the code and put 1 into the middle of a big square grid (25x25). To make it work out online, I changed the size ...


3

Yes, within 5 folds, the bear imprint can be made. The approach taken here is to draw the bear design on the paper and start folding in symmetrical parts to match all the points onto one another. Steps: 1.Fold the paper in half for symmetry. (fold-1) all other folds will be done on the folded paper. 2.align the most bottom point and the second bottom point ...


3

If we assume these are regular dice being rolled one at a time, the answer would be ...unless, of course,


3



3

I think the story is: Here's the image:


3

Assuming the ! is a factorial, there are because: Process for finding sets:


2

There are such numbers. Reasoning: The total list:


2



2

Sketch of a solution (?) for b=5. I'm showing the unfinished picture to invite some feedback. General idea: highway with u-turns: So does this solve everything? One thing to observe is that there are indeed attacks on this setup that can only be defeated by forward planning. If an adversary could decide the next digit without notice they could kill us:


2

Here is 19


1

For the numbers 2,4,6,8 one solution is (6-2)/4+8=9 For the numbers 4,5,6,8 one solution is 8/(6-4)+5=9


1

We may have ambiguity here, but I will try to answer both cases. Please note there could be more answers, just that I will list one for each case. 8, 6, 4, 2: 8, 6, 4, 5:


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