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# Tag Info

33

Explanation: Bonus:

2

With branch-and-bound algorithm wrote in Python, I found couple of solutions with the highest possible sum of side (inspired with trolley813's partial answer). One of them is here:

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Here is a solution with minimal side-sum: This was found by brute-force. The code below will traverse between the bounds as identified by trolley813 and prints a solution with each side-sum (42 & 54 take a while but the rest are much faster). The output solutions are in row-major order. from itertools import combinations, permutations def ...

0

The numbers in the circles on the long sides of the three triangles all add up to 43.

3

A partial answer based on both my own's answer to the original question and Steve's comments on it:

4

A solution (probably many more of them exist):

4

If I understood everything correctly, this solution is also acceptable. If not - it's still very interesting and simple one. It's valid for $n=0,1,2\dots,28$

10

There is no limit to this! The Green Tao theorem tells you that the sequence of prime numbers contains arbitrarily long arithmetic progressions. This means that using $a+b\cdot n$ you can get as many primes as you want for some $a,b$ and consecutive values of $n$. But the theorem does not tell you how to find $a,b$. The longest known such sequence can be ...

10

The function with rule produces distinct primes for $n$ up to $25$. For proof, see the third bullet point on this list of prime number records. It is valid for $x=0,1,...,23$, so I substitute $n=x+2$ so that the set of valid inputs begins at $2$. The function is clearly strictly increasing and so the primes must be distinct. It is apparently the longest ...

5

For $n = 2, 3, 4, 5$, the function produces distinct prime numbers:

0

An approximate alternative: $\frac{13}{3} \cdot (\frac{10}{3} + \frac{7}{3} \cdot \frac{4}{3}) = 27.96 \approx 28$

7

Perhaps there is a slight trick to this one

8

1

I’m going to guess:

3

Another solution is: The solutions correspond to:

2

An image representation of Gareth's answer

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After looking at @hexomino's answer I came up with a slightly different proof.

4

This mistake lies:

2

I claim that the long term average number of cars for $N$ starting cars will be To prove this we can proceed by induction

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(NB: table of calculations below is unspoilered. Smart people, feel free to edit.) Initial remarks N=3 Tedious calculations 4 CARS Back: (1/4*0/3+1/4*3/3+1/4*3/3+1/4*3/3) 2nd-back: (1/4*0/3+1/4*2/3+1/4*3/3+1/4*3/3) 2nd-front: (1/4*0/3+1/4*1/3+1/4*2/3+1/4*3/3) Front: 0 5 CARS Back: (1/5*0/4+1/5*4/4+1/5*4/4+1/5*4/4+1/5*4/4) 2nd-back: (1/5*...

4

He is singing about 1, 2. Freddy is drawing eyeballs. 2½, 3. One iris got bigger. 3½, 4. The other got bigger. 5, 6. Aura passes from and to irises.

4

How to get the answer: Hence, So, Still being edited...

1

My solution Note Here we go (numbers in parentheses indicate point count): Bonus:

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Per comments, this answer is invalid - it doesn't follow the "in order" rule. Current score: 50 Working: Bonus:

2

Instead of a cartesian plane, let's consider instead a... That means rotating $n$, $m$ and $k$ around, so that So armed with this nomenclature, let's run a few iterations starting with only the $(0,0)$ point: ...we can conclude that the number of moves needed to clear the... ...row/diagonal is...

6

It is possible for the $3\times3$ square: Here is an explanation of how I found those solutions: For $11\times11$:

5

5

Since we have I'm sure I remember a recent puzzle based on the same idea, but I'm failing to find it. There was Aha, found it: I'm not sure whether this should be regarded as a duplicate of that one. It isn't the same question but it's clearly closely related.

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Previous answer, valid before alteration to question. The maximum area is if one Note that

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Based on my previous answer of a similar question, I created a Python program to brute-force all the possible solutions. To calculate the number of possible solutions: That is a big number, but not too big for a computer with some time. However, there is two caveats: Here is the code: from dataclasses import dataclass from enum import Enum from typing ...

2

It seems a,b,c,d,e values are not unique; here is another answer;

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If

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The digits are out of order, but the sum still works. The given digits are on the base line. Operations: square, multiply, subtract, add, factorial.

5

4

An alternate approach:

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Additional solution, keeeping the digits in order:

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The train is travelling at: Working:

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Here is a solution that keeps the order of the digits:

11

Here's an attempt, since you allow squaring as an operation:

4

The medians have values ranging from 2 to 8, so exactly one of these values does not appear as the median of one of the 6 rows/columns. Here are three templates that allow you to choose any missing median value. For completeness, here are the solutions this produces:

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No 2: No 3: No 4: No 5:

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Here is one solution:

0

The number of ways of inserting brackets into $n$ objects is given by the $n−1$ Catalan number. https://en.wikipedia.org/wiki/Catalan_number.

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Partial answer with quick upper and lower bounds for the range. I believe the highest number you can get should be: The lowest should be: Which means that at most there could be: That said, the problem states numbers, which means we need to include decimals. Heading out shortly, but thoughts about an answer that probably won't actually help:

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$2020$

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Starting off short before the end of this answer we take

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My best solution so far is This is done with the following thermometer Here are three other thermometers with the same score: If you find any more thermometers that have this score and are not symmetric to any of the three in this post I would be happy to see them. Symmetry here means any combination of: Translation Rotation Mirroring Reversing the ...

2

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Here is one way to do it General Strategy

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