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021.py
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021.py
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"""
Project Euler Problem 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 collections
import math
def euler():
sum_of_amicable_numbers = 0
proper_divisor_sums = {}
for x in range(2, 10000):
divisor_sum = 1
for y in range(2, math.ceil(math.sqrt(x))):
if x % y == 0:
divisor_sum += y + x // y
proper_divisor_sums[x] = divisor_sum
for x, y in proper_divisor_sums.items():
if 1 < y < len(proper_divisor_sums) and x == proper_divisor_sums[y] and x != y:
sum_of_amicable_numbers += x
print(sum_of_amicable_numbers)
def compute():
# Compute sum of proper divisors for each number
divisorsum = [0] * 10000
for i in range(1, len(divisorsum)):
for j in range(i * 2, len(divisorsum), i):
divisorsum[j] += i
# Find all amicable pairs within range
ans = 0
for i in range(1, len(divisorsum)):
j = divisorsum[i]
if j != i and j < len(divisorsum) and divisorsum[j] == i:
ans += i
return str(ans)
euler()