Solution to Problem 21

Author: daniel-s-ingram

Project Euler/Problem 21/sol1.py

@@ -0,0 +1,42 @@
+#-.- coding: latin-1 -.-
+from __future__ import print_function
+from math import sqrt
+'''
+Amicable Numbers
+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.
+'''
+try:
+	xrange		#Python 2
+except NameError:
+	xrange = range	#Python 3
+
+def sum_of_divisors(n):
+	total = 0
+	for i in xrange(1, int(sqrt(n)+1)):
+		if n%i == 0 and i != sqrt(n):
+			total += i + n//i
+		elif i == sqrt(n):
+			total += i
+
+	return total-n
+
+sums = []
+total = 0
+
+for i in xrange(1, 10000):
+	n = sum_of_divisors(i)
+	
+	if n < len(sums):
+		if sums[n-1] == i:
+			total += n + i
+
+	sums.append(n)
+
+print(total)