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56 lines
1.2 KiB
Python
56 lines
1.2 KiB
Python
from math import sqrt
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"""
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Amicable Numbers
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Problem 21
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Let d(n) be defined as the sum of proper divisors of n (numbers less than n
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which divide evenly into n).
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If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and
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each of a and b are called amicable numbers.
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For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55
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and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and
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142; so d(284) = 220.
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Evaluate the sum of all the amicable numbers under 10000.
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"""
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def sum_of_divisors(n):
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total = 0
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for i in range(1, int(sqrt(n) + 1)):
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if n % i == 0 and i != sqrt(n):
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total += i + n // i
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elif i == sqrt(n):
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total += i
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return total - n
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def solution(n):
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"""Returns the sum of all the amicable numbers under n.
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>>> solution(10000)
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31626
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>>> solution(5000)
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8442
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>>> solution(1000)
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504
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>>> solution(100)
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0
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>>> solution(50)
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0
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"""
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total = sum(
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[
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i
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for i in range(1, n)
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if sum_of_divisors(sum_of_divisors(i)) == i and sum_of_divisors(i) != i
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]
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)
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return total
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if __name__ == "__main__":
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print(solution(int(str(input()).strip())))
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