Hi Haskellers, I was solving this problem from project euler to study haskell. I came up whit the following solution and I was wondering if there is a more optimized and concise solution.
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 Thanks in advance ggarcia _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
