Help with try and open 6.txt file and print as perfect or not

[stullis]

The problem is likely on line 13. Since sum1 is a global variable, it cannot be reset inside a function unless that function uses the keyword "global". In any case, global functions should not be used like that. Here's a rewrite of the code. The commented out section should work for you and lines 28 through 33 are a test for 6:

import math

KB = 1024
MB = 1024 * 1024

def perfect_number(number, factors):
    sum_of_factors = sum(factors) - number
    if sum_of_factors == number:
        return f"The number {number} is a Perfect number!"
    else:
        return f"The number {number} is not a Perfect number!"

def factor_number(n):
    root = int(round(math.sqrt(n), 0))
    return [(x, n // x) for x in range(1, root + 1) if n % x == 0]

if __name__ == '__main__':
    """with open('6.txt', buffering=1*MB) as bigggg:
        for num in bigggg:
            number = int(num)
            factors = []

            for pair in factor_number(number):
                factors.extend(pair)

            print(perfect_number(number, factors))"""

    factors = []

    for pair in factor_number(6):
        factors.extend(pair)

    print(perfect_number(6, factors))

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You may want to make factors into a set instead of list using my code. Otherwise, square numbers may not be correctly evaluated.