13

Here is a relatively quick way to get an approximate answer. Let $k$ be the number of steps to the end (in this case $12$), and $n$ the number of sides on the die. First you need to look at the end game. Now to estimate the number of throws until we reach the end game: We need to minimise this.


11

It is Lower bound: Upper bound:


11

Aunt Isabel received Reasoning After looking at the numbers Therefore the totals are, by child's age


10

With a little more work than Jaap's very nice approximate answer we can explicitly calculate the values we need. Pick a die -- say it has $n$ sides -- and let $a_j$ be the expected number of turns when you are $j$ spaces away from winning. Obviously $a_0=0$, and Jaap's insight about how the game ends shows that $a_j=1/n$ when $1\leq j\leq n$. After that we ...


10

This game is also known as Treblecross. There is a neat way to think about this game. As soon as you place an X next to another X, or one away from another X you lose because the opponent can make three in a row. Therefore the Xs need to be a distance of at least 2 apart. You can think of each X being in the middle of a 3x1 bar that you place on the board, ...


8

Here is a quick way to do it


7

Simulating in C# / LINQPad: void Main() { testIt(); } Random rnd = new Random(); // Define other methods and classes here void testIt() { var tries = new List<List<int>>(); for (int n = 2; n <= 12; ++n) { var dieTries = new List<int>(); tries.Add(dieTries); for (var t = 0; t < 500000; ++t) ...


7

@JimN has shown it is doable in 12. What remains to be done is Proof of optimality:


7

I found several by hand with 13 steps, then wrote a program... which found hundreds more solutions with 13 steps, and I was beginning to think that maybe there was no solution for 12 and then my program spat-out this: Edit1: Another solution found by the silly program: Edit2: I have my code available. It is written in Java. I tested it against the earlier ...


7

Here's another solution using the fact that $\triangle ABC$ and $\triangle ABD$ have the same area if $AB \parallel CD$: The reason is that $CE = DF$, so $\frac12 AB \cdot CE = \frac12 AB \cdot DF$. Now, since $DG \parallel EK$ and $BD \parallel EG$, we have $\triangle DGK = \triangle DEG = \triangle BEG$ in area: The result is hence


6

The diameter of the smallest circle is The configuration is an instance of a modified by the geometric transformation We start with a similar but more symmetrical figure Then we apply inversion in the circle centred at R It is easy to verify that, by the properties of inversion in a circle, circle E is mapped to line OJ (circle E passes through R so it ...


6

The diameter of the smallest circle is Let's put the figure in a Cartesian plane: (half of the figure is omitted due to symmetry) Denote the radii of the four circles by $R_P = \frac12$, $R_A$, $R_B$, and $R_C$, respectively. We see that $R_A$ is the ordinate of $A$ and that $1 - R_B$ is the ordinate of $B$. Since $PA = R_A + \frac12$ is also the distance ...


5

Old school proof of optimality:


4

Another solution, with smallest possible maximum entry subject to minimizing the sum: If you ignore the sum, the smallest possible maximum entry is smaller by $1$:


3

I'll get things started with


3

Minimum amount of boxes on display I haven't found a proven minimum for question 2 yet but the lowest I've gotten so far is With display box sizes of


3

I would be surprised if this question hasn't appeared before so apologies in advance if answering a duplicate We'll do question 2 first This means the answer to question 1 is And question 3


3

This is actually much easier than it looks. You can simply use any consistent method for separating the the tetrad pieces. By consistent I mean that if two stage 3 positions can be brought to a stage 4 position by the same set of moves, then the two positions get the same tetrad twist number. Note however that you don't need to figure out what those moves ...


2

For completeness, here is my model solution which is mostly identical to @Edward Doolittle's but uses an ever so slightly more convenient map.


2

Multiple answers seem to be valid, for example It seems that the question was in fact missing a tag, as I needed to make use of to arrive at the answer process which was: Perhaps the flavour text about the donuts Taking the clues given: For the "wrong" answer given: Correct answers for example (explaining the examples given at the start of ...


2

I decided to let my computer solve it, and it found the following solution: This is the same as hexomino's latest solution, who updated his post with this shortly before I posted this answer. Assuming my program is correct, this is optimal. To show this is indeed a solution: The second-best solution has box sizes that are coprime: Using 4 or 5 display ...


2

I found this: If this is right, it took me four tries, so while not trivial, I'm not sure I'd label it as 'Expert' unless that is low on your scale of difficulty. Otherwise maybe I am not understanding the rules. But I can see how the mechanics could make later puzzles much more interesting. Keep up the good work!


2

Yes, it's possible:


2



2

Here's a "manual" solution. Let's start by noticing that there are 21 changes between consecutive digits (there are implicit zeroes to the left and the right), and an operation can eliminate at most 2 of them, one at either end of the changed digit string. This means that 11 moves is an absolute lower bound. So, let's start by finding all the ...


2

Just write with a common denominator and you will get $$\frac{x-17}{(x-9)(x-11)(x-13)(x-14)}=0,$$ which has solution $x=17$.


2

Here is one such tiling:


1

Tada! No, sorry, I actually cheated.


1

Answer Reasoning


1

Partial answer (a crude upper bound): Because


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