46 lines
1.1 KiB
Python
46 lines
1.1 KiB
Python
#!/usr/bin/env python
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"""
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Created on 15 Sep 2018
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 021 of Project Euler
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https://projecteuler.net/problem=21
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"""
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from utils import timeit
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def sum_divisors(n):
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return sum(i for i in range(1, n // 2 + 1) if n % i == 0)
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@timeit("Problem 021")
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def compute():
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"""
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Let d(n) be defined as the sum of proper divisors of n (numbers
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less than n 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
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pair and 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,
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44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are
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1, 2, 4, 71 and 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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n = 10_000
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ans = 0
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for i in range(1, n):
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value = sum_divisors(i)
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if i != value and sum_divisors(value) == i:
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ans += i
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return ans
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if __name__ == "__main__":
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print(f"Result for Problem 021: {compute()}")
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