39 lines
1.2 KiB
Python
39 lines
1.2 KiB
Python
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##########################################################################
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# Let d(n) be defined as the sum of proper divisors of n
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# (numbers 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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#
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# For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110;
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# therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
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#
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# Evaluate the sum of all the amicable numbers under 10000.
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##########################################################################
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def sumDivisors(x):
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divisor_sum = 0
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for j in range(1, int(x / 2) + 2):
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if (x % j) is 0:
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divisor_sum += j
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return divisor_sum
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amicable_number = []
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for i in range(1, 10000):
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if i in amicable_number:
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continue
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testing_number = sumDivisors(i)
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if (i == sumDivisors(testing_number)) and (testing_number != i):
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amicable_number.append(i)
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if testing_number not in amicable_number:
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amicable_number.append(testing_number)
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sum_of_numbers = sum(amicable_number)
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print(sum_of_numbers)
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# Solution: 31626
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