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First solution in Clear category for Perfect Number Checking by mortonfox
from collections import Counter
from itertools import product
from functools import reduce
from operator import mul
def prime_factors(n):
i = 2
while i * i <= n:
while n % i == 0:
yield i
n //= i
i += 1
if n > 1: yield n
def divisors(n):
factors = Counter(prime_factors(n))
powers = [[factor ** i for i in range(count + 1)] for factor, count in factors.items()]
for prime_power_combo in product(*powers):
yield reduce(mul, prime_power_combo, 1)
def is_perfect(n: int) -> bool:
return sum(divisors(n)) == 2 * n
print("Example:")
print(is_perfect(3))
# These "asserts" are used for self-checking
assert is_perfect(6) == True
assert is_perfect(2) == False
assert is_perfect(28) == True
assert is_perfect(20) == False
assert is_perfect(496) == True
assert is_perfect(30) == False
assert is_perfect(8128) == True
assert is_perfect(100) == False
assert is_perfect(500) == False
assert is_perfect(1000) == False
print("The mission is done! Click 'Check Solution' to earn rewards!")
Oct. 11, 2023
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