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Anonymous Thu 05 Dec 03:15:13 2019 No.11200408 View Reply Original Report

Quoted By: >>11200454 >>11200681 >>11200687

You have to guess a number between 900,000 and 1,000,000. The fourth digit is 7 and the last digit is 6. Therefore, the number would be 9**7*6. Is there a mathematical way to list all the numbers, or maybe a program I can write? Thanks in advance

Anonymous

Anonymous Thu 05 Dec 2019 03:23:36 No.11200426 Report

Quoted By:

934796

Anonymous

Anonymous Thu 05 Dec 2019 03:29:47 No.11200436 Report

Quoted By: >>11200466 >>11200531

911716
911726
911736
911746
...
999796

there i did it

Anonymous

Anonymous Thu 05 Dec 2019 03:34:07 No.11200454 Report

Quoted By: >>11200464 >>11200466

>>11200408
Since you are only changing three numbers then you would need to find all the 3-permutations of a set of 9 elements. Thus the answer would be 9x8x7 = 504 possible options

Anonymous

Anonymous Thu 05 Dec 2019 03:37:51 No.11200464 Report

Quoted By: >>11200493

>>11200454
not OP but is there some place a beginner and non math savvy person can read about calculating total permutations?
also would it not be from a set of 10 elements?

Anonymous

Anonymous Thu 05 Dec 2019 03:38:10 No.11200466 Report

Quoted By: >>11200493

>>11200454
God I hope you’re joking.
>>11200436
This anon has the right idea but seems to have forgotten that 0 is a digit. This is a simple for loop or 3 nested if you’re lazy.

Anonymous

Anonymous Thu 05 Dec 2019 03:46:42 No.11200493 Report

Quoted By: >>11200502 >>11200726

>>11200464
>>11200466
shit who do i believe

Anonymous

Anonymous Thu 05 Dec 2019 03:48:13 No.11200502 Report

Quoted By:

>>11200493
Unless you forgot to specify that the digits needed to be different, there’s no reason to worry about permutations.

Anonymous

Anonymous Thu 05 Dec 2019 03:59:53 No.11200531 Report

Quoted By:

>>11200436

this but including 0. the number of different number of this format is 1000.

Anonymous

Anonymous Thu 05 Dec 2019 05:07:12 No.11200681 Report

Quoted By:

>>11200408
>:

> shitty python implementation

from itertools import permutations

def insert_digits(n):
num_list = list(n)
num_list.insert(0, 9)
num_list.insert(4, 6)
num_list.insert(3, 7)

num = int(''.join(map(str, num_list)))
return(num)

perm_digits = list(permutations(range(0, 9), 3))
perm_nums = [insert_digits(x) for x in perm_digits]
print(perm_nums)

Anonymous

Anonymous Thu 05 Dec 2019 05:09:42 No.11200687 Report

Quoted By:

>>11200408
> shitty python implementation

https://pastebin.com/JyqtqWVu

Anonymous

Anonymous Thu 05 Dec 2019 05:32:27 No.11200726 Report

Quoted By:

>>11200493
hes right, easy to do it with cartesian product
> shitty python implementation:
https://pastebin.com/igRrFcAH

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