New answers tagged combinatorics
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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