Solution to problem 12

src/Python/Problem012.py

@@ -0,0 +1,67 @@
+#!/usr/bin/env python3
+"""
+Created on 01 Jan 2018
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for problem 12 of Project Euler
+https://projecteuler.net/problem=12
+"""
+
+import itertools
+from utils import timeit
+from math import sqrt, floor
+
+
+# Returns the number of integers in the range [1, n] that divide n.
+def num_divisors(n):
+    end = floor(sqrt(n))
+    divs = []
+    for i in range(1, end + 1):
+        if n % i == 0:
+            divs.append(i)
+    if end**2 == n:
+        divs.pop()
+    return 2*len(divs)
+
+
+
+@timeit("Problem 12")
+def compute():
+    """
+    The sequence of triangle numbers is generated by adding the natural
+    numbers. So the 7th triangle number would be:
+    1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
+
+    The first ten terms would be:
+    1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+
+    Let us list the factors of the first seven triangle numbers:
+
+     1: 1
+     3: 1,3
+     6: 1,2,3,6
+    10: 1,2,5,10
+    15: 1,3,5,15
+    21: 1,3,7,21
+    28: 1,2,4,7,14,28
+
+    We can see that 28 is the first triangle number to have over five divisors.
+
+    What is the value of the first triangle number to have over five hundred
+    divisors?
+    """
+
+    triangle = 0
+    for i in itertools.count(1):
+        # This is the ith triangle number, i.e. num = 1 + 2 + ... + i =
+        # = i*(i+1)/2
+        triangle += i
+        if num_divisors(triangle) > 500:
+            return str(triangle)
+
+
+if __name__ == "__main__":
+
+    print(f"Result for Problem 12: {compute()}")