20

It looks like you start and then and


12

I'll use my own box labelling so that the method is shorter to describe. Step 1: First I'll remove the labels, then pick any three boxes and test them. Whichever of the three boxes is the median I'll label B, and the other two boxes are A and C. Step 2: Step 3: Reasoning: Proof of optimality:


10

I think this works: The adjacent number pairs are restricted to where each pair sums to Note that it is necessary to because


7

The following works: Confession: Someone asked in comments whether there was a logical path to the solution. Not really, especially as part of the actual path to the solution was getting a nudge from OP's comment on someone else's answer! But I can say a few things: Of course, if you wanted to be really methodical you could just ask a computer. There are ...


7

It can be done in and that is optimal. Optimality: Schedule:


7

I'll answer this in the form of a poem: Just follow the instructions, and you'll have the answer. All credit to:


7

It is to divide the cake into 9 equal pieces in this way. Details: Some more details: And


6

Here is another one: Note that for any permutation of the first 4 columns there are 6 matching permutations of the last 4 that give rise to another solution. And similar for rows. So this is actually a family of solutions.


6

Here is a solution inspired by Beastly Gerbil's observation in a comment that: From here, we can finish with almost no casework: We didn't even use the condition in the third column, though we can check that it's satisfied at the end.


5

As I commented already, I have a heuristic argument that because Based on this belief, I quickly coded a program in Factor: : good-number? ( k n -- ? ) [ swap mod 0 = ] [ nip prime? ] 2bi or ; : next-numbers ( k seq -- k+1 seq' ) [ 1 + ] [ [ 10 * 10 <iota> [ + ] with map ] map concat ] bi* [ drop ] [ [ good-number? ] with filter ] 2bi ; 1 9 [...


5

The number $9$ Option 1: Option 2: So the full solution is


5

There is indeed a simple solution. The key thing to realize is


5

The statement is Proof: Having some equal items does


4

The ratio is Proof


4

Glorfindel solved this in a few minutes, but for your entertainment I would like to show the "solution" as a Python script. Download the prime number file from https://primes.utm.edu/lists/small/millions/ Note that the code could be optimized. It updates the figure for 1 million steps in about a minute on my pc. (sorry, can't wrap the code in ...


4

This is an improvement to Alemin's strategy that makes the maximum number of questions... Because


4

Here's another way to solve this puzzle. We can actually proceed one digit at a time, and the middle column product trick (see below) seems way too neat to be a coincidence. Step 1: Where can the 1 go? It can't be in the rightmost column (it's too small to be subtracted, and if it's added, the other two digits in the column would have to be equal to each ...


4

The answer is because, for example, if T2 has side-lengths then a triangle T1 such as More generally, we can consider T2 with side-lengths and T1 with side-lengths


4

Let's do the second part first: The cylinder will now have exactly 2/3 of water left, because Continuing from here, we can get the one third by or by


3

The volumes of the boxes from left to right are I found it easier to do the second task first.


3

partial answer: further insight:


3

Intresting question! Just to fix an upper bound: Because:


3

I think we need at least with the example coloring of To see this is the minimum, first consider Then, consider one of the colors used to color a center cell: Therefore, we conclude that


3

The smallest $n$ will be And here’s the grid (all possible grids will be rotations of this): This is because Now to find the grid


3

We can achieve with the following pattern: This is optimal because: (In fact, this pattern can tile the plane - there's no need to restrict it to a 5×5 grid.)


2

Case-bashing here is not so bad (and perhaps somewhat necessary since some of the examples come very close to working). For ease, I will use the following notation $$\begin{array}{ccccc} & & A1 & A2 & A3 \\ & & \times & B1 & B2 \\ \hline & C1 & C2 & C3 & C4 \\D1 & D2 & D3 & ...


2

Here is an example


2

First, we have to find the areas of the pieces (the inner square $I$, the midsections $M$, and the corners $C$) in terms of $r$ and the setting $2w$:


2

It seems like the parameters of the questions are as follows Under these conditions it looks like the minimum number of moves we need is Reasoning


2

I'm not sure how I'd prove this is optimal, but my answer is: Reasoning:


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