# Tag Info

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

Well, I guess we are talking about the:

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

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

3

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

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.

2

Anyone is welcome to edit and improve this answer, so don't checkmark this answer. Bolded numbers mean they need to be solved or are impossible. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

7

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 ...

3

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.

1

For sake of completeness, I have done a computer search and have found that 40 $\tau$s is the absolute minimum one can achieve. This can be done using Ian's expressions for 1–13 and 17–20, KSmarts' expression for 15, and two new expressions for 14 and 16: The only expressions requiring 3 $\tau$s are those for 11, 19, and 20, and exhaustive search has shown ...

5

7

Additional solution, keeeping the digits in order:

7

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

11

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

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.

0

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:

1

$2020$

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