New answers tagged mathematics
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Are these colored sets closed under multiplication?
Additional observation/stuff that can be useful to expand the answer:
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Alice splits the bill not too generously with Bob
Alice must choose a number with no less than
The lowest such number is
Bob's strategy is:
If we do this, then the score Bob can earn is defined by the following recurrence relation:
We can ...
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Alice splits the bill not too generously with Bob
So Alice chooses from numbers where Bob can on average guess more than 60% of the digits, but as little more than 60% as possible. Alice should chose numbers with
When she tells Bob how many 1's ...
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Are these colored sets closed under multiplication?
Problem statement and headline result
This is a partial answer to the tougher question asked in comments on msh210's answer:
(3bis) How can the real numbers be partitioned in two sets that are both ...
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Fill this Sudoku variant so that the sums of numbers in the outlined regions are all different
Unique solution:
Reasoning:
To refer to cells easier, let the top row be cells A1-A5, and label further cells accordingly.
To start, notice that there are 9 groups consisting of at most 2 cells, with ...
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What is the expected time this will take?
Analysis
Let
$$
\begin{aligned}
W = & \text{ number of white balls}\\
B = & \text{ number of black balls}\\
N = & \text{ }W + B\\
\end{aligned}
$$
At each step, there are three possibile ...
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Are these colored sets closed under multiplication?
Question 1: Is it necessarily true that at least one of the sets is closed under multiplication?
Question 2: Is it necessarily true that both sets are closed under multiplication?
Question 3: Is it ...
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Starting with 2014 "+" signs and 2015 "−" signs, you delete signs until one remains. What’s left?
Initially, there are an odd number of - signs. Neither of the two permitted operations can change the fact that there are an odd number of ...
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Escape from the magic prison
After looking at the problem, I think this problem resembles the "expectancy" problem in mathematics. And this is my way of solving it:
3 + 9 + 27 = 39.
You need at least 39 moves to ...
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What is the expected time this will take?
I’m not giving an exact answer but a thinking process:
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What is the expected time this will take?
Extremely partial answer:
Define f(n,m) as the expected number of moves until all balls are the same color. Therefore we get the following two equations due to basic probability -
f(0,m) = f(n,0) = 0 ...
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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:
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Starting with 2014 "+" signs and 2015 "−" signs, you delete signs until one remains. What’s left?
From My knowledge and persective as
"+""+"="+"
"+""-"="-"
"-""-"="+"
as there are 2014 "+" and ...
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Guesstimate a multiple choice exam
As you know 5 ans are true, only 15 quetion ae left.
Now True left = 14-5 = 9 , False left = 6
Now you want to maximize your score, so leave it on GOD and randomly select any 9 quetion as True and ...
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Starting with 2014 "+" signs and 2015 "−" signs, you delete signs until one remains. What’s left?
Answer: -
No matter how many + symbols there are, they can all be reduced to a single + by ...
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Escape from the magic prison
Upper Bound = 36
These were found using a Depth-First Search with pruning.
$36$ presses:
...
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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:
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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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Everyday words for terminology
Since no one has answered I will submit my answer, but something doesn't quite fit at the end so I don't think I'm fully correct.
Like other commenters, I first saw
I didn't see this, but ...
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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:
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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 $n×n$ ...
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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 ...
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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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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 +...
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Getting lost on a Circular Track
For the non-zero segment size:
For the zero segment size case:
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What is the least number of colours Peter could use to color the 3x3 square?
I got
by "coloring" with numbers:
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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 ...
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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 ...
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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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What is the least number of colours Peter could use to color the 3x3 square?
The minimum is
because
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Starting with 2014 "+" signs and 2015 "−" signs, you delete signs until one remains. What’s left?
Or without a brain:
remove 1007 pairs of "-" : you now have 2014+1007=3021 "+" and a single "-"
recursively remove "+" by coupling them with the only "-&...
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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 ...
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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 &...
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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 ...
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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 ...
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What is the optimal number of function evaluations?
I believe the minimum is
Strategy
Explicit:
Example
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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 ...
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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 ...
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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
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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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Unorthodox angle measuring device
The device is a:
Then:
This device also fits the three hints:
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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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Unorthodox angle measuring device
Same as SquareFinder. I have a rule but no matching device. Maybe it inspires someone.
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