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:
Only top voted, non community-wiki answers of a minimum length are eligible
