31
votes
Accepted
Starting with 2014 "+" signs and 2015 "−" signs, you delete signs until one remains. What’s left?
Notice that, if you set + = 0 and - = 1, then the mod 2 of the sum does not change under either transformation. Hence, the final sum must be equivalent to the original sum, 2015, mod 2. Therefore the ...
- 335
22
votes
Accepted
What is the least number of colours Peter could use to color the 3x3 square?
The minimum is
because
- 28.2k
17
votes
Starting with 2014 "+" signs and 2015 "−" signs, you delete signs until one remains. What’s left?
Now that an officially new user has answered let me remark that + and - cry out to be read as
and the entire thing can ...
- 7,735
16
votes
What is the least number of colours Peter could use to color the 3x3 square?
Basically a beginner here.
Start with a diagonal. All three cells must have unique colours:
Then, the two unshaded corners must be given unique colours because both of them have a diagonal with the ...
- 524
13
votes
Starting with 2014 "+" signs and 2015 "−" signs, you delete signs until one remains. What’s left?
I think this answer is equivalent to BlazingSnow's answer, but instead of assigning numbers to the symbols and then reducing mod 2, I do a case analysis and observe parity of the count of each symbol:
...
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13
votes
Accepted
Unorthodox angle measuring device
The device is a:
Then:
This device also fits the three hints:
- 18.1k
12
votes
Accepted
Getting lost on a Circular Track
For the non-zero segment size:
For the zero segment size case:
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11
votes
Escape from the magic prison
The best upper bound I've found is 39. Here are a few sequences which should solve every possible configuration:
...
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11
votes
Accepted
10
votes
Escape from the magic prison
Upper Bound = 36
The best result was found using a Depth-First Search with pruning.
$36$ presses:
111233233311312321212131213231313222
$37$ presses:
...
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8
votes
Getting lost on a Circular Track
For the single-point exit, you can't do better than @fljx's answer.
Watch out, because:
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8
votes
What is the optimal number of function evaluations?
I will ignore the fact that the function is convex. I suspect that this fact doesn't help the worst-case performance.
Definition
Let $x_{answer}$ be the value of $x$ which for $f(x)$ is minimal.
Game ...
- 2,143
7
votes
Accepted
Manhattan distance
The place to start is:
Now let us place:
Moving down the chain:
And the next:
You guessed it:
Finishing up:
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7
votes
The subtraction game
This is (surprisingly) actually a win for Alice!
If she chooses [9, 5, 4, 11, 6, 14, 3, 8, 15, 12, 18, 7, 16, 24, 13, 36, 63, 48] she is guaranteed to win for any number greater than 96.
Using the ...
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7
votes
What is the least number of colours Peter could use to color the 3x3 square?
As described in many answers, five colors is the minimum. Here we bring in the theory of pandiagonal Latin squares to show some hidden features of the solution and allow a generalization to $n×n$ ...
- 1,080
6
votes
Starting with 2014 "+" signs and 2015 "−" signs, you delete signs until one remains. What’s left?
We can delete any two symbols so order doesn't matter, it suffices to track the counts: let p, m be the number of plus and minus symbols.
We have three kinds of ...
- 61
5
votes
5
votes
What is the optimal number of function evaluations?
I think the answer is:
Order of queries:
Let's analyze this below. Let $1\le x_m\le 200$ be such that $f(x_m)$ is minimum.
The only way to tell if $f(x_m)$ is minimum is that:
Also, if we currently ...
- 958
5
votes
Accepted
Combination lock on a pentagonal rotating table
My procedure is:
For the proof, first some setup ignoring the rotations:
The main claim:
Now to address the rotations:
To prove the claim,
Other thoughts:
- 166
5
votes
Accepted
4
votes
Manhattan distance
Since someone beat me to it, I won't post the solve path, but only the key observation, that
The solve is easiest if you
and the full solution is
- 2,172
4
votes
What is the least number of colours Peter could use to color the 3x3 square?
I got
by "coloring" with numbers:
- 221
3
votes
Starting with 2014 "+" signs and 2015 "−" signs, you delete signs until one remains. What’s left?
Straightforward thinking:
There are three ways to remove two signs:
Remove two +s, then add one back (effectively removing one +...
- 3,196
3
votes
What is the optimal number of function evaluations?
I will provide an easy approach (but I don’t think this is the optimized one).
After discussion with @thisIs4d I realized that my approach is wrong, but I don't have a better way to improve it for ...
- 735
2
votes
Escape from the magic prison
I've reworked my code entirely in an attempt to find an optimum via A* search. I'm done encoding the worlds and simulation of moves, but I've yet to find a suitable heuristic.
My current heuristic is ...
- 3,278
2
votes
Unorthodox angle measuring device
I'm not really sure about the precision/accuracy but I think this could be on the right track. If not the correct method, then at least that it's related to the
If we convert the first term in each ...
- 38.9k
2
votes
Visual puzzle involving mathematical operations
I’m not sure if there exists some kind of coincidence, but I noticed…
step 1. add up the numbers with the same color of each row/ column
$$
\begin{array}{cc|ccccc}
row/col & value & orange &...
- 735
2
votes
Everyday words for terminology
Here's my attempt. I think a few are shaky, but it's plausible.
First group:
Second group:
Third group:
And finally:
- 7,280
2
votes
Guesstimate a multiple choice exam
Same result as @fandango96 but perhaps a simple more generic explanation.
What matters is:
Then optimal is:
because this way:
OP mentions tricky perhaps because:
1
vote
Only top scored, non community-wiki answers of a minimum length are eligible