Solved problem 9

0008-sum-1000-digit-number.py

@@ -46,7 +46,9 @@ def traditional():
 84580156166097919133875499200524063689912560717606\
 05886116467109405077541002256983155200055935729725\
 71636269561882670428252483600823257530420752963450"
+
     l = []
+
     for i in range(len(number) - 5):
         l.append(int(number[i]) * int(number[i+1]) * \
         int(number[i+2]) * int(number[i+3]) * int(number[i+4]))

0009-pythagorean-triplet.py

@@ -0,0 +1,29 @@
+"""
+Problem 9
+
+A Pythagorean triplet is a set of three natural numbers, a  b  c, for which,
+
+a2 + b2 = c2
+For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
+
+There exists exactly one Pythagorean triplet for which a + b + c = 1000.
+Find the product abc.
+"""
+import math
+
+def is_perfect_square(n):
+    return int(math.sqrt(n))**2 == n
+
+def pythagorean():
+    # Building a list of perfect squares
+    l = [x for x in range(2, 1000)]
+    ans = 0
+
+    for a in l:
+        for b in l:
+                c = 1000 - (a + b)
+                if a**2 + b**2 - c**2  == 0:
+                    ans = a*b*c
+    return ans
+
+print "Answer by traditional method:", pythagorean()