New answers tagged

10



3

When I wrote the question I had a more concrete solution in mind. The other answers given are very computational or mathematical. They don't really show how the solution is constructed imho. So here is what I had in mind.


3

Let's start by filling out the gaps in the grid with letters to make it easier to describe: .----.---.---.---. | 5 | y | 3 | x | :----+---+---+---: | a | | | c | :----+---+---+---: | b | | | d | :----+---+---+---: | 12 | p | q | 6 | '----'---'---'---'


3

If the baskets are named with the rows and columns: A B C D E F G H I J K L M N O P I will first try and swap the baskets' containers where the two will be solved with just 1 swap: Then finish the puzzle with the smallest number of swaps: So the smallest number of swaps needed is EDIT: <-> is a swap, e.g. A<->B is basically swapping whatever'...


4

Let us consider the series https://oeis.org/A005428 We will now show by induction that the following three statements are all true: Note that wherever I say "set" below I should really have said "multi set" but that would have been tedious so I didn't. All three statements are easily verified for small $n$. Now observe that by ...


5

You are looking for a partition of $40$ with the minimum number of parts that is a common refinement of all $154$ partitions of $40$ into at most $3$ parts. You can satisfy all $154$ scenarios with Note that no container can contain more than 14 boxes because the centers might demand 14, 13, and 13, respectively. Update: I had initially assumed that the ...


17

This puzzle could have almost have been given the tag, though that may have given a big hint. You can think of each rectangle you pick as a move, Here is a proof for why this is the minimal number of rectangles: (Thanks to loopywalt for plugging the final gap in the proof)


14

Answer: Strategy: Visualisation: Proof:


9

The answer is Yes.


6

Say we have $b$ boys and $g$ girls, then we have $${g\choose 2} = {3\over 10}{b+g\choose 2}$$ so we have this equation $$3b^2 +3b(2g-1)= 7g(g-1)$$ Now for each $g$ find $b$. If we put $x = 2(b+g)-1$ and $y= 2g-1$ we get $$\boxed{10y^2-7=3x^2}$$ Here are some values


6

I'm not sure if this is the only possible solution to the puzzle, but I found that if: Again, there may be more solutions based on the rules outlined in the puzzle.


2

Matt Parker hosted a challenge to solve this puzzle last year. The participants found that there were two distinct solutions for 6x6, one for 7x7, and none for 8x8 or 9x9 (up to rotations/reflections). https://youtu.be/G0i_YSFvMb0?t=366 According to comments on the video, the search was extended with no solutions found up to 14x14, and you can prove that no ...


11

I don't know how you would be able to deduce a solution to this, so I used a computer. It turns out that the solution is unique (ignoring rotation and reflection). My C# program does a straightforward exhaustive search: using System; using System.Collections.Generic; namespace TempProg { class PSEGolombSquare { private const int N ...


11

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, ...


3

Yes, it's possible:


2



0

Following is a partial answer and I am hoping that someone can help me prove that it works. Let us label the cups from 1 to 13. Before the game begins, the magician and the apprentice write down all the (13*12)/2 = 78 combinations that I can choose, on a blackboard. To number 1, they allocate, any arbitrary 4 numbers: a,b,c and d. The only things they make ...


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

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 ...


-1

If the store is a stationery or office supply shop, then the following possibilities are valid for the number of boxes and the number of pencils they contain. (The $b$s represent the boxes and the numbers represent the corresponding numbers of pencils.) $b_1=5 \times 2^0$ $b_1‹b_2 \le5\times2^1$ $b_2‹b_3\le5\times2^2$ $b_3‹b_4\le5\times2^3$ $b_4‹b_5\le5\...


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


6

There is one solution for $G$, which is as follows Notice first that Now Suppose instead


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