Solution to problem 21

src/Python/Problem021.py

@@ -0,0 +1,42 @@
+#!/usr/bin/env python3
+"""
+Created on 15 Sep 2018
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for problem 21 of Project Euler
+https://projecteuler.net/problem=21
+"""
+
+from utils import timeit
+
+def sum_divisors(n):
+    return sum(i for i in range(1, n//2+1) if n%i==0)
+
+@timeit("Problem 21")
+def compute():
+    """
+    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.
+    """
+    n = 10000
+    sum_amicable = 0
+    for i in range(1, n):
+        value = sum_divisors(i)
+        if i != value and sum_divisors(value) == i:
+            sum_amicable += i
+    return sum_amicable
+
+
+if __name__ == "__main__":
+
+    print(f"Result for Problem 21: {compute()}")