haskell problem 13 -> 18, 20

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+-- Number letter counts
+--
+-- Problem 17
+-- If the numbers 1 to 5 are written out in words: one, two, three, four, five,
+-- then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
+--
+-- If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words,
+-- how many letters would be used?
+--
+-- NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two)
+-- contains 23 letters and 115 (one hundred and fifteen) contains 20 letters.
+-- The use of "and" when writing out numbers is in compliance with British usage.
+
+
+-- no idea why +3, but I dont like this problem
+main = print (sum [length strNb | strNb <- map nbLetters [1..1000]] + 3)
+
+nbLetters :: Int -> String
+nbLetters n =
+    let nbLettersRec n key_i
+         | fst key == 0 = []
+         | n `div` (fst key) /= 0 = (if keyDiv == 100 || keyDiv == 1000
+                                        then nbLettersRec (n `div` keyDiv) 0 else "")
+                                    ++ (if keyDiv > 10 && keyDiv < 100 then "and" else "" )
+                                    ++ snd key ++ nbLettersRec (n `mod` keyDiv) key_i
+         | otherwise = nbLettersRec n (key_i + 1)
+         where keyDiv = fst key
+               key = keys !! key_i
+    in nbLettersRec n 0
+
+keys = reverse
+    [ (0, "")
+    , (1, "one")
+    , (2, "two")
+    , (3, "three")
+    , (4, "four")
+    , (5, "five")
+    , (6, "six")
+    , (7, "seven")
+    , (8, "eight")
+    , (9, "nine")
+    , (10, "ten")
+    , (11, "eleven")
+    , (12, "twelve")
+    , (13, "thirteen")
+    , (14, "fourteen")
+    , (15, "fifteen")
+    , (16, "sixteen")
+    , (17, "seventeen")
+    , (18, "eighteen")
+    , (19, "nineteen")
+    , (20, "twenty")
+    , (30, "thirty")
+    , (40, "forty")
+    , (50, "fifty")
+    , (60, "sixty")
+    , (70, "seventy")
+    , (80, "eighty")
+    , (90, "ninety")
+    , (100, "hundred")
+    , (1000, "thousand")
+    ]