I wrote a computer program and it showed that $18$ moves is the optimum.
Here is one such solution:
Oddly enough, even if you relax the condition of alternating white and black moves, it cannot be done in fewer moves.
For $3\times3$ the optimal number of moves is $16$.
Without the need to alternate moves the optimum is $14$ moves, for example just by ...
(Kind of) analytical solution that only requires small amount of calculations, (potentially) doable by hand.
First step: we can safely drop 2, 3 and 7 from the equation as those digits are used in 23 and 17. Now, we need to build a prime from: 5, 11, 13, 17, 19, 23, 29 and 31.
Second step: let's try to build the shortest number possible from these numbers. ...
The answer is 7, It's a simple extension of the Magic Calculator trick.
Each card represents a bit, with the upper left index value being the value of that bit.
To perform it as a trick, you ask the volunteer to pick a number between 1 and 64, and hand you the cards on which their number appears. Add up the index values on the selected cards, and you have ...
I managed to find the 8 in
So we are trying to find the 5th largest card in a bunch of 12, by measuring them in batches of four. Here's my strategy:
Now we have identified, for certain, a couple of cards we can exclude: A1 and A2 (both have at least 8 cards smaller than them), and B4, C2, C3 and C4 (all have at least 5 cards bigger than them). We also have ...
Here is a solution with
The colouring of the empty squares ($2\times2$ in white, $4\times4$ in yellow, $6\times6$ in orange, $8\times8$ in red) shows how the pattern generalises to any $2k\times2k$ board.
For $8\times 8$ this solution is optimal.
I can ensure that I receive at least
To get this amount of coins, I could apply the following strategy:
My friend can also ensure that I do not get more coins than this in the following way:
So it appears that I should have chosen a fairer game to distribute the coins...
Between 10 friends, there are only 45 unique pairs.
Each friend will participate exactly 9 times, which means after 45 rounds everyone will be level 10.
Since it's impossible to get an item on the first & last run, there is a theoretical maximum of 43 items you can get.
However, due to some restrictions, it seems to me like it's impossible to get this ...
This is similar to another answer but with a smaller area. I worked it out completely independently, then noticed its similarity. The 3" dish is in a different place, and it is not an adjustment based on that answer. It was generated by a C program I wrote for this purpose. It gave my previous answers and has been spitting out smaller ...
Thinking about this problem in reverse is easier, i.e. think about the trees that need to remain. Find the fewest number of trees that can be arranged in five lines with 4 trees in each line. To minimise that number, we need as many trees as possible to do double duty by being part of multiple lines.