🔒 Lucky card

[knotty patrol]

There are n numbers of cards labeled from 1 to n, each with an index from 1 to n counting from left to right, and the index changes as the number of cards reduces. To find the lucky card, do as follows:

1. Keep the cards whose index modulus 3 is equal to 2 and discard all others.
2. If there is more than 1 card left, repeat 1., else return the number labeled on the last card.
   Limitation: 1 second of run time and 1023 MB of memory.
   Input: n varies from 1 to 10^9

[hybrid olive]

@knotty patrol

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[pure latch]

What’s your code so far?

[knotty patrol]

I haven't coded it yet due to work issues but here's my formula (sorry for the late reply):

[pure latch]

Main thing here is to turn it into code. And then we can help you when you get stuck. But you gotta get stuck first 😎

[knotty patrol]

Finally, I can sit back and code, while I'm coding it, can I get a validation on this formula?

[knotty patrol]

Here's my code, I'd like to hear your suggestions:

import math

n = int(input("enter the amount of cards: "))

result = 2 + sum(3**x for x in range(1, math.floor(math.log(n - (2 + sum(3**x1 for x1 in range(1, math.floor(math.log(n, 3))))), 3)) + 1))

print(result)

[hybrid olive]

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