Add Haskell solution for Problem 12

Haskell/ProjectEuler.hs

@@ -2,9 +2,11 @@ module ProjectEuler
 ( isPrime
 , digitSum
 , sumProperDivisors
+, countDivisors
 ) where
 
 import Data.Char (digitToInt)
+import Data.List (nub)
 
 isPrime :: (Integral n) => n -> Bool
 isPrime 1 = False
@@ -22,3 +24,9 @@ digitSum n = sum $ map digitToInt $ show n
 
 sumProperDivisors :: (Integral a) => a -> a
 sumProperDivisors n = sum [ if x /= y then x + y  else x | x <- [2..ceiling $ sqrt $ fromIntegral n], let y = n `div` x, n `mod` x == 0 ] + 1
+
+
+countDivisors :: (Integral a) => a -> Int
+countDivisors n = length $ nub $ concat [ [x, n `div` x] | x <- [1..limit], n `mod` x == 0 ]
+    where limit = floor $ sqrt $ fromIntegral n
+

Haskell/p012.hs

@@ -0,0 +1,33 @@
+-- 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?
+
+import Data.List (nub)
+
+import ProjectEuler (countDivisors)
+
+triangNumbers :: [Int]
+triangNumbers = scanl1 (+) [1..]
+
+triang500 :: Int
+triang500 = head [ x | x <- triangNumbers, countDivisors x > 500 ]
+
+main = do
+    let result = triang500
+    putStrLn $ "Project Euler, Problem 12\n"
+        ++ "Answer: " ++ (show result)