48
votes
Accepted
43
votes
Accepted
30
votes
Accepted
Do Langford squares exist?
Langford squares are not possible.
Consider the middle two rows of a $2n \times 2n$ Langford square. They, like all rows, must contain $n$'s. Any $n$ must have a partner $n$ in its column that's $n+1$ ...
- 26.3k
25
votes
Accepted
Selectively neglected collection
These mannequins all have something in common...
So what we need to do to assemble the remaining mannequins is too...
And as tmpearce points out the comments,
- 6,683
25
votes
Accepted
21
votes
Accepted
20
votes
Accepted
16
votes
15
votes
15
votes
Accepted
What is the number of ways to spell French word « chrysanthème »?
The word can be spelled
First we
Then we
If we
- 18.3k
15
votes
Accepted
13
votes
The shorter the message, the larger the prize (version II)
I will give a very weak lower bound (improved based on xnor's idea), and a minor improvement to user1502040's answer.
Lower bound:
Proof:
11111111111111111111 ...
- 2,976
12
votes
7 mathematicians around the clock in prison
HEADS UP: If you find this post interesting in principle but too terse or messy for your liking @Andrew Savinykh has written a more friendly version.
UPDATE: Here is a program demoing the strategy
...
- 19.1k
11
votes
Accepted
7 mathematicians around the clock in prison
This is the same solution by loopy-walt, presented a bit differently and adding some context. It seemed to me that the solution has some leaps that may be obvious to many people but also could be ...
- 597
11
votes
Accepted
11
votes
10
votes
Accepted
9
votes
Accepted
Clock hands get it Right
Here is a way I like to think about it
Where does your reasoning fall down
- 133k
9
votes
Accepted
Do non-trivial Skolem squares exist?
No non-trivial Skolem squares exist.
First, observe that any number $k$ appearing in a Skolem square must be part of an axis-aligned square of width and height $k$ whose four vertices are all $k$. For ...
- 26.3k
9
votes
9
votes
5 chess pieces dominating a 5x5 grid
Came up with an alternative solution from the other answers
- 301
9
votes
16 queens puzzle
For the question of maximizing the number of sets of $8$ queens, define a graph with a node for each of the $92$ solutions for $8$ queens and an edge for each pair of solutions that share a cell. Now ...
- 12k
8
votes
Accepted
Save now! All the digits at half the price
Let's define some terms:
Now we can begin.
Let's break the problem down now.
Now for the final count!
- 8,317
8
votes
Two arcs equal three arcs
Here's one solution. See this album for larger images. The multiset sum of the two arcs on the left, when placed on top of each other, equals the multiset sum of the three arcs on the right when ...
- 511
8
votes
Playing Mastermind against an angel and the devil
I don't know if it's optimal, but the devil can show at least
I determined this by
- 25.7k
7
votes
Accepted
7
votes
Accepted
n*n*n Rubik's cube algorithm
For a cube of edge n:
n=1 is trivial.
n=3 can be solved with known algorithms, such as the CFOP method.
n=2 has faster algorithms but can also be solved as just the corners of the n=3 case.
For n&...
- 76k
7
votes
Accepted
7
votes
Accepted
Visiting all strings by swapping
By introducing a dummy node that is adjacent to all other nodes, we have a traveling salesman problem on a graph with $37$ nodes and $130$ edges. The distance between nodes is $0$ if either node is ...
- 12k
7
votes
Accepted
Pay each amount with at most two coins
An easy lower bound is
The minimum turns out to be
You can solve the problem via integer linear programming as follows. Let $C=\{1,\dots,100\}$ be the set of coins. Let $P=\{(i,j): 1 \le i \le j \...
- 12k
Only top scored, non community-wiki answers of a minimum length are eligible