First kata - find the sum all multiples of 3 or 5 less than 1000

Sixth solution.

A whole new approach was attempted here. We have removed the loop throughout every single number to check if it is a multiple of 3 ou 5 to a mathematical approach using 2 different formulaes.
1. the sum of the multiples 3 or 5 = sum of multiples of 3 + sum of multiples of 5 - sum of multiples of 3 * 5
2. the sum of multiples of 3 = ((n * (n + 1)) / 2) * d
    where n = 1000 / 3 (upper limit divided by the multiple)
                d = multiple (in this case 3)

with this solution we solve the performance issue of the previous solution by providing a constant-time solution.
previous solution = 1 operation per number (i.e 10^10 will cause the function to run 10^10 steps)
new solution = 3 operation per limit (1 to compute the sum of multiples of 3, 1 to compute the sum of multiples of 5, 1 to compute the sum of multiples of 3 * 5)

haskell/katas/01/FirstKata.hs

@@ -2,10 +2,5 @@ module FirstKata (
   fkSum
 ) where
 
-fkSum :: Int -> [Int] -> Int
-fkSum a [] = 0
-fkSum a (b) = foldl (+) 0 (filter (\x -> isMultiple x b) [1..a])
-
-isMultiple :: Int -> [Int] -> Bool
-isMultiple a [] = False
-isMultiple a (x:xs) = if (mod a x == 0) then True else isMultiple a xs
+fkSum u x y = f x + f y - f (x * y)
+ where f d = let n = u `div` d in (n * (n + 1) `div` 2) * d

haskell/test/01/FirstKataTest.hs

@@ -1,7 +1,7 @@
 import Test.HUnit
 import FirstKata 
 
-testFkSumFirst = TestCase $ assertEqual "result should be 23" 23 (fkSum 9 [3,5])
-testFkSumSecond = TestCase $ assertEqual "result should be 233168" 233168 (fkSum 999 [3,5])
+testFkSumFirst = TestCase $ assertEqual "result should be 23" 23 (fkSum 9 3 5)
+testFkSumSecond = TestCase $ assertEqual "result should be 233168" 233168 (fkSum 999 3 5)
 
 suite = TestList [ testFkSumFirst, testFkSumSecond ]