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1

There are My process: At this point, The numbers, and divisors are:


0

Testing several bottles together instead of letting a rat drink from the first, second etc bottle until it dies or survives all the bottles never gains more information, so it is pointless. Now to find the number of bottles: Assume the king who never lies tells you that he picked one bottle, poisoned the other 99, and then either poisoned the last bottle or ...


0

Some ground work: $\begin{matrix}(7) & 720 \\ (6) & 840 \\ (5) & 504 \\ (5,2) & 504 \\ (4) & 210 \\ (4,3) & 420 \\ (4,2) & 630 \\ (3) & 70 \\ (3,3) & 280 \\ (3,2) & 420 \\ (3,2,2) & 210 \\ (2) & 21 \\ (2,2) & 105 \\ (2,2,2) & 105 \\ () & 1 \end{matrix}$ $\begin{matrix} (o3,o2) & 2520 \\ (o2,o2,...


2

Looking at the benefit of mixing... My gut feeling is


5

What is the minimum number of rats required to conclusively identify the value of K? The answer is 1 rat. It just might happen to not always be the maximum number required.


11

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1

If I... That would be a very inefficient solution to the problem, unless... So it's kind of a...


8

If I... ...then... ...unless...


4

I confirm @AlexeyBurdin's count of 470 solutions, which I obtained via integer linear programming as follows. Let $S=\{2, 4, 5, 6, 7, 8, 10, 11, 12, 13\}$ be the set of desired sums. Let binary decision variable $p_{x,y}$ indicate whether there is a point with coordinates $(x,y)$, let binary decision variable $r_x$ indicate whether row $x$ contains any ...


13

I'm pretty sure the minimum is If we consider the worst case scenario:


3

Here is a solution with "one component":


4

Doesn't look anywhere near unique to me: For example, or or or for a slightly different class of solution and probably more


7

I hope I understood correctly. Solution 1: Second solution Explanation:


7

The second rule states: "For each point there is exactly one other point so that their x-coordinates or their y-coordinates match". So we should be able to pair each point. We have an odd number of points to place, so ps:sorry I can't comment yet.


5

Pictorial analysis There are $27$ different possibilities to enter into the lock. Let's say two of them are connected if they have two common digits (in the same positions); so every possibility is connected to exactly $6$ others. We can show them on a diagram like this: We can also see here that two triple-digit possibilities which are cycled versions of ...


7

I think the best we can do is The following tries will cover all possibilities Proof that this is the best possible Alternative Proof (courtesy of Jaap Scherphuis in the comments) Bonus Not totally sure but the best I've managed to do is As follows


4

This is a standard application of the Burnside Lemma. I'll solve the more general case of a square with $n$ colours.


5

I think the answer is Counting


5

I have found with this grid I used a hill climbing algorithm that changes one value at a time. It accepts a move if it increases (or equals) the score, otherwise it rejects it. After all possible changes have been tried, it adds some random mutations and restarts the process. I ran multiple processes of this method for about a week and it only found this ...


9

Edit: Here is a new and improved answer of As follows


15

I used integer linear programming as follows. Let $C=\{1,\dots,6\}^2$ be the set of cells, and let $D=\{0,\dots,9\}$ be the set of digits. Let $P=\{(i_1,j_1,i_2,j_2,i_3,j_3)\}$ be the set of paths of length three ($|P|=1460$), and let $T \subseteq \{(d_1,d_2,d_3)\in D^3: d_1 \not= 0\}$ be the set of digit triples to be covered. (The one- and two-digit ...


5

As of now I have: I may be able to squeeze a few more


4

No guarantees of optimality, but I'll start us off with a score of


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