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Problem Euler #21

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.


import time
from math import sqrt
ts = time.clock()

def get_div(num):
    for n in xrange(2, int(sqrt(num)) + 1):
            if num % n == 0:
                yield n
                yield num/n

def get_sum(divisors):
    return sum([n for n in divisors]) + 1

if __name__ == '__main__':
    nums = [n for n in range(1, 10000) if n == get_sum(get_div(get_sum(get_div(n)))) \
            and n != get_sum(get_div(n))]
    print sum(nums)
    print time.clock() - ts
Categorie:Project Euler, python
  1. settembre 23, 2014 alle 2:29 pm

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