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Problem021.py
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Problem021.py
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"""
Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.
"""
from math import sqrt, floor
from time import time
def get_divisors(x):
divisors = {1}
for i in [i for i in range(2, floor(sqrt(x))) if x % i == 0]:
divisors.update([i, x//i])
return divisors
result = 0
N = 10_000
start_time = time()
for i in range(1, N + 1):
target = sum(get_divisors(i))
if i < target and i == sum(get_divisors(target)):
result+= i+target
final_time = time()
print(result)