13

Started by scaling the goal, I put one-by-one with some trial-and-error in the end to find this solution:


12

I think: Here's my proof:


7

It can be: If:


6

Here is one way to do it General Strategy


6

It is possible for the $3\times3$ square: Here is an explanation of how I found those solutions: For $11\times11$:


5

Having a Blokus Trigon set at hand was a big help! I arrived at a different actually the same solution as pointed out: My approach was rather straightforward, here's how I went around it: Hint 1: Hint 2:


4

The medians have values ranging from 2 to 8, so exactly one of these values does not appear as the median of one of the 6 rows/columns. Here are three templates that allow you to choose any missing median value. For completeness, here are the solutions this produces:


4

Here is one solution: I solved it by


4

Here is one solution:


4

Starting off short before the end of this answer we take


4

I asked a question over on Math.SE, pointing to this puzzle, and @Elaqqad gave a very long answer with lots of links to math that might help. @Elaqqad also produced a concrete solution with only 59 tests — beating @noedne's long-standing solution of 63 tests! Background math The "wolves and sheep" puzzle is a specific case of non-adaptive group testing. ...


3

We can test this statement by the following: However I doubt this is the "lateral thinking" the OP has in mind.


3

You can't make it in a fair way, as JS1 answered. But given lateral thinking tag - you could have dice where one influences the other. Perhaps by magic or magnets. For example, first die is fair (1/6 chance of any number) and when it rolls 1, there is 6/11 chance of rolling 1 on the second one and 1/11 to roll any other number. Symmetry tells us 6 should ...


2

From JS1's answer we already see that this is impossible for normal dice. But assuming the dice's result are dependent on each other as if they could communicate, This is how it works: Let's pick a "nice" solution: Actual numbers:


2

I am not giving a new solution. But I'd like to propose a nicer expression of the solution given for the general case $n=2^k$. In fact my solution is the same as noedne's. And the explanation why it works is the same as OHO's.


2

It is Because


2

No 2: No 3: No 4: No 5:


1

Previous answer, valid before alteration to question. The maximum area is if one Note that


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