New answers tagged combinatorics
1
I think there are
Counting
I would remove these counters
3
The solution to this problem and its generalizations (multiple numbers on larger grids) can be found in this integer sequence:
https://oeis.org/A302980
You can see the actual solutions here:
https://oeis.org/A302980/a302980.txt
17
Although the puzzle is most likely to be solved without a computer, and we already have a winner, here are all 16 solutions, just for the record:
There are some symmetries in there, of course. Whether rotations and flips should count as "different boards" is a matter of interpretation.
Found with the following (quick and dirty) Java program that jus ...
6
Here is another way of completing this puzzle:
4
It feels like there's more slack in here for other solutions
1.
2.
3.
35
Assuming standard Minesweeper rules, here’s one solution (with $ X $ = a mine):
EDIT: In response to Euphoric in the comments, I solved this purely by logical deduction with a bit of educated guessing to make things easier on me. But if you really want to know how I did it, here’s a rigorous solution:
4
Yes: the solution was already mentioned in this answer, sourced to this page.
2
Here's the solution:
It's interesting to note that the domination problem has very few solutions for 3 queens on a 6x6 chessboard as compared to other possibilities. (Neither of those links contains the answer! Just background info.)
Perhaps the key realisation is that
Before that, starting with
0
The question doesn't refer to the splitting, but the fragments themselves, which makes the answer
because
4
The answer is
Reasoning
3
The smallest achievable sum is
Achieved as follows, one possibility
Reasoning
Number of Solutions
1
Answer
example 1
example 2
example 3
2
I ran a computer program to find all solutions. Armed with the answer, I've tried to construct an argument for why those are the only solutions, but have not quite managed that.
Firstly, I'll assume without proper proof that:
Now those remaining numbers need to be arranged so that
This leaves the following solutions:
I have not found any proof for the ...
2
Deleted answer (misunderstood instruction)
4
The answer is
First, we calculate the number of blendings where the $e$'s and $c$'s are labelled ($e_1$, $e_2$,...). Then we will count the number of such words where two $e$'s or two $c$'s are next to each other and subtract it accordingly.
Now
Similarly
Finally
Wait a minute.
1
No, it's not possible:
If you start by weighing 3 vs 3 and it comes out equal, you only have two weighings to find the fake coin (of unknown weight) among six.
In weighing 2, if you have more than three potential fakes on the scale, and it comes out unbalanced, then the remaining weighing (with only three potential results) will not allow you to pick the ...
0
I'm not entirely sure how to single out the exact ball here as the description given seems to imply that I have four groups of three balls and only three times to use the scale in order to determine the ball.
Based on that, I can use standard logic to figure out which group the different ball is in. Let's say that our groups are labeled A, B, C, and D, and ...
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